# Solve hanging lantern problem

(Redirected from Lantern Problem)
Solve hanging lantern problem is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

There are some columns of lanterns hanging from the ceiling. If you remove the lanterns one at a time, at each step removing the bottommost lantern from one column, how many legal sequences will let you take all of the lanterns down?

For example, there are some lanterns hanging like this:

```🏮 🏮 🏮
🏮 🏮
🏮
```

If we number the lanterns like so:

```1 2 4
3 5
6
```

You can take like this: [6,3,5,2,4,1] or [3,1,6,5,2,4]
But not like this: [6,3,2,4,5,1], because at that time 5 is under 4.

In total, there are 60 ways to take them down.

Input:
First an integer (n): the number of columns.
Then n integers: the number of lanterns in each column.
Output:
An integer: the number of sequences.

For example, the input of the example above could be:

```3
1
2
3
```

And the output is:

```60
```

Output all the sequences using this format:

```[1,2,3,…]
[2,1,3,…]
……
```

Related

## ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32
Translation of: Python – Math sample
```BEGIN # find the number of ways of taking down columns of lanterns          #
# translation of the Python "Math" sample                               #

PR precision 256 PR            # set number of digits for LONG LONG INT #
# may need to be increased, depending on the number of lanterns #

OP   FACTORIAL = ( INT n )LONG LONG INT:
BEGIN
LONG LONG INT result := 1;
LONG LONG INT long n := 0;
TO n DO result *:= ( long n +:= 1 ) OD;
result
END # FACTORIAL # ;

PROC get lantern = ( REF[]INT arr )LONG LONG INT:
BEGIN
LONG LONG INT res := 0;
INT           tot := 0;
FOR i FROM LWB arr TO UPB arr DO
tot +:= arr[ i ]
OD;
res := FACTORIAL tot;
FOR i FROM LWB arr TO UPB arr DO
res OVERAB FACTORIAL arr[ i ]
OD;
res
END # get lantern # ;

BEGIN
INT n;
print( ( "Columns> " ) );
read( ( n, newline ) );
[ 1 : n ]INT a;
print( ( "lanterns per column> " ) );
FOR i TO n DO
read( ( a[ i ] ) )
OD;
print( ( whole( get lantern( a ), 0 ), newline ) )
END
END```
Output:
```Columns> 3
lanterns per column> 1 2 3
60
```
```Columns> 5
lanterns per column> 1 2 3 4 5
37837800
```
```Columns> 8
lanterns per column> 1 2 3 4 5 6 7 8
73566121315513295589120000
```

## APL

Translation of: Pascal
```lanterns ← { (!+/⍵) ÷ ×/!⍵ }
```
Output:
```      lanterns 1 2 3
60
lanterns 1 3 3
140
```

Of course, for the simple sequences from 1, we can use iota to generate them instead of typing them out:

```      lanterns ⍳3 ⍝ same as lanterns 1 2 3
60
lanterns ⍳4
12600
lanterns ⍳5
37837800```

## BASIC

### BASIC256

Translation of: FreeBASIC

The result for n >= 5 is slow to emerge

```arraybase 1
n = 4
dim a(n)
for i = 1 to a[?]
a[i] = i
print "[ ";
for j = 1 to i
print a[j]; " ";
next j
print "] = "; getLantern(a)
next i
end

function getLantern(arr)
res = 0
for i = 1 to arr[?]
if arr[i] <> 0 then
arr[i] -= 1
res += getLantern(arr)
arr[i] += 1
end if
next i
if res = 0 then res = 1
return res
end function```
Output:
`Same as FreeBASIC entry.`

### Commodore BASIC

Translation of: Python

The (1,2,3) example takes about 30 seconds to run on a stock C64; (1,2,3,4) takes about an hour and 40 minutes. Even on a 64 equipped with a 20MHz SuperCPU it takes about 5 minutes.

```100 PRINT CHR\$(147);CHR\$(18);"***     HANGING LANTERN PROBLEM      ***"
110 INPUT "HOW MANY COLUMNS "; N
120 DIM NL(N-1):T=0
130 FOR I=0 TO N-1
140 : PRINT "HOW MANY LANTERNS IN COLUMN"I+1;
150 : INPUT NL(I):T=T+NL(I)
160 NEXT I
170 DIM I(T),R(T)
180 SP=0
190 GOSUB 300
200 PRINT R(0)
220 END
300 R(SP)=0
310 I(SP)=0
320 IF I(SP)=N THEN 420
330 IF NL(I(SP))=0 THEN 400
340 NL(I(SP))=NL(I(SP))-1
350 SP=SP+1
360 GOSUB 300
370 SP=SP-1
370 R(SP)=R(SP)+R(SP+1)
390 NL(I(SP))=NL(I(SP))+1
400 I(SP)=I(SP)+1
410 GOTO 320
420 IF R(SP)=0 THEN R(SP)=1
430 RETURN
```
Output:
```***     HANGING LANTERN PROBLEM      ***

HOW MANY COLUMNS ? 4
HOW MANY LANTERNS IN COLUMN 1 ? 1
HOW MANY LANTERNS IN COLUMN 2 ? 2
HOW MANY LANTERNS IN COLUMN 3 ? 3
HOW MANY LANTERNS IN COLUMN 4 ? 4
12600```

### FreeBASIC

Translation of: Python
```Function getLantern(arr() As Uinteger) As Ulong
Dim As Ulong res = 0
For i As Ulong = 1 To Ubound(arr)
If arr(i) <> 0 Then
arr(i) -= 1
res += getLantern(arr())
arr(i) += 1
End If
Next i
If res = 0 Then res = 1
Return res
End Function

Dim As Uinteger n = 5
Dim As Uinteger a(n)
'Dim As Integer a(6) = {1,2,3,4,5,6}
For i As Ulong = 1 To Ubound(a)
a(i) = i
Print "[ ";
For j As Ulong = 1 To i
Print a(j); " ";
Next j
Print "] = "; getLantern(a())
Next i
Sleep```
Output:
```[ 1 ] = 1
[ 1 2 ] = 3
[ 1 2 3 ] = 60
[ 1 2 3 4 ] = 12600
[ 1 2 3 4 5 ] = 37837800```

### QBasic

Works with: QBasic version 1.1
Works with: QuickBasic version 4.5
Translation of: FreeBASIC

The result for n >= 5 is slow to emerge

```FUNCTION getLantern (arr())
res = 0
FOR i = 1 TO UBOUND(arr)
IF arr(i) <> 0 THEN
arr(i) = arr(i) - 1
res = res + getLantern(arr())
arr(i) = arr(i) + 1
END IF
NEXT i
IF res = 0 THEN res = 1
getLantern = res
END FUNCTION

n = 4
DIM a(n)
FOR i = 1 TO UBOUND(a)
a(i) = i
PRINT "[";
FOR j = 1 TO i
PRINT a(j); " ";
NEXT j
PRINT "] = "; getLantern(a())
NEXT i
END
```
Output:
`Same as FreeBASIC entry.`

### PureBasic

Translation of: FreeBASIC

The result for n >= 5 is slow to emerge

```;;The result For n >= 5 is slow To emerge
Procedure getLantern(Array arr(1))
res.l = 0
For i.l = 1 To ArraySize(arr(),1)
If arr(i) <> 0
arr(i) - 1
res + getLantern(arr())
arr(i) + 1
EndIf
Next i
If res = 0
res = 1
EndIf
ProcedureReturn  res
EndProcedure

OpenConsole()
n.i = 4
Dim a.i(n)
For i.l = 1 To ArraySize(a())
a(i) = i
Print("[")
For j.l = 1 To i
Print(Str(a(j)) + " ")
Next j
PrintN("] = " + Str(getLantern(a())))
Next i
Input()
CloseConsole()```
Output:
`Same as FreeBASIC entry.`

See Visual Basic

### Visual Basic

Works with: Visual Basic version 6

Note: Integer may overflow if the input number is too big. To solve this problem, simply change Integer to Long or Variant for Decimal.

#### Recursive version

Main code
```Dim n As Integer, c As Integer
Dim a() As Integer

Private Sub Command1_Click()
Dim res As Integer
If c < n Then Label3.Caption = "Please input completely.": Exit Sub
res = getLantern(a())
Label3.Caption = "Result：" + Str(res)
End Sub

Private Sub Text1_Change()
If Val(Text1.Text) <> 0 Then
n = Val(Text1.Text)
ReDim a(1 To n) As Integer
End If
End Sub

Private Sub Text2_KeyPress(KeyAscii As Integer)
If KeyAscii = Asc(vbCr) Then
If Val(Text2.Text) = 0 Then Exit Sub
c = c + 1
If c > n Then Exit Sub
a(c) = Val(Text2.Text)
Text2.Text = ""
End If
End Sub

Function getLantern(arr() As Integer) As Integer
Dim res As Integer, i As Integer
For i = 1 To n
If arr(i) <> 0 Then
arr(i) = arr(i) - 1
res = res + getLantern(arr())
arr(i) = arr(i) + 1
End If
Next i
If res = 0 Then res = 1
getLantern = res
End Function```
Form code
```VERSION 5.00
Begin VB.Form Form1
Caption         =   "Get Lantern"
ClientHeight    =   4410
ClientLeft      =   120
ClientTop       =   465
ClientWidth     =   6150
ScaleHeight     =   4410
ScaleWidth      =   6150
StartUpPosition =   3
Begin VB.CommandButton Command1
Caption         =   "Start"
Height          =   495
Left            =   2040
TabIndex        =   5
Top             =   3000
Width           =   1935
End
Begin VB.ListBox List1
Height          =   1320
Left            =   360
TabIndex        =   4
Top             =   1440
Width           =   5175
End
Begin VB.TextBox Text2
Height          =   855
Left            =   3360
TabIndex        =   1
Top             =   480
Width           =   2175
End
Begin VB.TextBox Text1
Height          =   855
Left            =   360
TabIndex        =   0
Top             =   480
Width           =   2175
End
Begin VB.Label Label3
Height          =   495
Left            =   2040
TabIndex        =   6
Top             =   3720
Width           =   2295
End
Begin VB.Label Label2
Caption         =   "Number Each"
Height          =   375
Left            =   3960
TabIndex        =   3
Top             =   120
Width           =   1695
End
Begin VB.Label Label1
Caption         =   "Total"
Height          =   255
Left            =   960
TabIndex        =   2
Top             =   120
Width           =   1455
End
End
Attribute VB_Name = "Form1"
Attribute VB_GlobalNameSpace = False
Attribute VB_Creatable = False
Attribute VB_PredeclaredId = True
Attribute VB_Exposed = False```

#### Math solution

Translation of: Python

Reimplemented "getLantern" function above

```Function getLantern(arr() As Integer) As Integer
Dim tot As Integer, res As Integer
Dim i As Integer
For i = 1 To n
tot = tot + arr(i)
Next i
res = factorial(tot)
For i = 1 To n
res = res / factorial(arr(i))
Next i
getLantern = res
End Function

Function factorial(num As Integer) As Integer
Dim i As Integer
factorial = 1
For i = 2 To n
factorial = factorial * i
Next i
End Function```

### Yabasic

Translation of: FreeBASIC

The result for n >= 5 is slow to emerge

```n = 4
dim a(n)
for i = 1 to arraysize(a(),1)
a(i) = i
print "[ ";
for j = 1 to i
print a(j), " ";
next j
print "] = ", getLantern(a())
next i

sub getLantern(arr())
local res, i
res = 0
for i = 1 to arraysize(arr(),1)
if arr(i) <> 0 then
arr(i) = arr(i) - 1
res = res + getLantern(arr())
arr(i) = arr(i) + 1
fi
next i
if res = 0  res = 1
return res
end sub```
Output:
`Same as FreeBASIC entry.`

## FutureBasic

```_elements = 5

local fn GetLantern( arr(_elements) as long ) as long
long i, res = 0
for i = 1 to _elements
if arr(i) != 0
arr(i) = arr(i) - 1
res = res + fn GetLantern( arr(0) )
arr(i) = arr(i) + 1
end if
next
if res = 0 then res = 1
end fn = res

long i, j, a(_elements)
for i = 1 to _elements
a(i) = i
print "[";
for j = 1 to i
if j == i then print a(j); else print a(j); ",";
next
print "] = "; fn GetLantern( a(0) )
next

HandleEvents```
Output:
```[1] = 1
[1,2] = 3
[1,2,3] = 60
[1,2,3,4] = 12600
[1,2,3,4,5] = 37837800
```

## J

Translation of APL:

```lanterns=: {{ (!+/y) % */!y }}<
```

Example use:

```   lanterns 1 2 3
60
lanterns 1 3 3
140
```

Also, a pedantic version where we must manually count how many values we are providing the computer:

```pedantic=: {{
assert. ({. = #@}.) y
lanterns }.y
}}
```

And, in the spirit of providing unnecessary but perhaps pleasant (for some) overhead, we'll throw in an unnecessary comma between this count and the relevant values:

```   pedantic  3, 1 2 3
60
pedantic  3, 1 3 3
140
```

If we wanted to impose even more overhead, we could insist that the numbers be read from a file where tabs, spaces and newlines are all treated equivalently. For that, we must specify the file name and implement some parsing:

```yetmoreoverhead=: {{
pedantic ({.~ 1+{.) _ ". rplc&(TAB,' ',LF,' ') fread y
}}
```

Examples of this approach are left as an exercise for the user (note: do not use commas with this version, unless you modify the code to treat them as whitespace).

Finally, enumerating solutions might be approached recursively:

```showlanterns=: {{
arrange=. (\$ \$ (* +/\)@,) y \$&>1
echo 'lantern ids:'
echo rplc&(' 0';'  ')"1 ' ',.":|:arrange
echo ''
cols=. <@-.&0"1 arrange
recur=: <@{{
todo=. (#~ ~:&a:) y -.L:0 x
if. #todo do.
next=. {:@> todo
,x <@,S:0 every next recur todo
else.
<x
end.
}}"0 1
echo 'all lantern removal sequences:'
echo >a:-.~ -.&0 each;0 recur cols
}}
```

Example use:

```   showlanterns 1 2 1
lantern ids:
1 2 4
3

all lantern removal sequences:
1 3 2 4
1 3 4 2
1 4 3 2
3 1 2 4
3 1 4 2
3 2 1 4
3 2 4 1
3 4 1 2
3 4 2 1
4 1 3 2
4 3 1 2
4 3 2 1
```

## jq

The main focus of this entry is illustrating how cacheing can be added to the naive recursive algorithm. Some trivial optimizations are also included.

With these changes, the algorithm becomes quite performant. For example, the C implementation of jq accurately computes the value for the lantern configuration [1,2,3,4,5,6,7] in less than a second on a 2.53GHz machine.

For lantern configurations with more than 2^53 permutations, the accuracy of the C implementation of jq is insufficient, but the Go implementation (gojq) can be used. For the configuration [1,2,3,4,5,6,7,8], gojq takes just over 4 minutes to produce the correct answer on the same machine.

```# Input: an array representing a configuration of one or more lanterns.
# Output: the number of distinct ways to lower them.
def lanterns:

def organize: map(select(. > 0)) | sort;

# input and output: {cache, count}
def n(\$array):
(\$array | organize) as \$organized
|  (\$organized|length) as \$length
| if   \$length == 1 then .count = 1
elif \$length == 2 and \$organized[0] == 1 then .count = (\$organized | add)
else .cache[\$organized|tostring] as \$n
| if \$n then .count = \$n
else reduce range(0; \$length) as \$i ({cache, count: 0, a : \$organized};
.a[\$i] += -1
| .a as \$new
| n(\$new) as {count: \$count, cache: \$cache}
| .count += \$count
| .cache = (\$cache | .[\$new | tostring] = \$count)
| .a[\$i] += 1 )
| {cache, count}
end
end;
. as \$a | null | n(\$a) | .count;

"Lantern configuration => number of permutations",
([1,3,3],
[100,2],
(range(2; 10) as \$nlanterns
| [range(1; \$nlanterns)])
| "\(.) => \(lanterns)" )```

Invocation

```gojq -n -rf lanterns.jq
```
Output:
```Lantern configuration => number of permutations
[1,3,3] => 140
[100,2] => 5151
[1] => 1
[1,2] => 3
[1,2,3] => 60
[1,2,3,4] => 12600
[1,2,3,4,5] => 37837800
[1,2,3,4,5,6] => 2053230379200
[1,2,3,4,5,6,7] => 2431106898187968000
[1,2,3,4,5,6,7,8] => 73566121315513295589120000
```

## Julia

```""" rosettacode.org /wiki/Lantern_Problem """

using Combinatorics

function lanternproblem(verbose = true)
println("Input number of columns, then column heights in sequence:")
inputs = [parse(Int, i) for i in split(readline(), r"\s+")]
n = popfirst!(inputs)
println("\nThere are ", multinomial(BigInt.(inputs)...), " ways to take these ", n, " columns down.")

if verbose
idx, fullmat = 0, zeros(Int, n, maximum(n))
for col in 1:size(fullmat, 2), row in 1:size(fullmat, 1)
if inputs[col] >= row
fullmat[row, col] = (idx += 1)
end
end
show(stdout, "text/plain", map(n -> n > 0 ? "\$n " : "  ", fullmat))
println("\n")
takedownways = unique(permutations(reduce(vcat, [fill(i, m) for (i, m) in enumerate(inputs)])))
for way in takedownways
print("[")
mat = copy(fullmat)
for (i, col) in enumerate(way)
row = findlast(>(0), @view mat[:, col])
print(mat[row, col], i == length(way) ? "]\n" : ", ")
mat[row, col] = 0
end
end
end
end

lanternproblem()
lanternproblem()
lanternproblem(false)
```
Output:
```Input number of columns, then column heights in sequence:
3 1 2 3

There are 60 ways to take these 3 columns down.
3×3 Matrix{String}:
"1 "  "2 "  "4 "
"  "  "3 "  "5 "
"  "  "  "  "6 "

[1, 3, 2, 6, 5, 4]
[1, 3, 6, 2, 5, 4]
[1, 3, 6, 5, 2, 4]
[1, 3, 6, 5, 4, 2]
[1, 6, 3, 2, 5, 4]
[1, 6, 3, 5, 2, 4]
[1, 6, 3, 5, 4, 2]
[1, 6, 5, 3, 2, 4]
[1, 6, 5, 3, 4, 2]
[1, 6, 5, 4, 3, 2]
[3, 1, 2, 6, 5, 4]
[3, 1, 6, 2, 5, 4]
[3, 1, 6, 5, 2, 4]
[3, 1, 6, 5, 4, 2]
[3, 2, 1, 6, 5, 4]
[3, 2, 6, 1, 5, 4]
[3, 2, 6, 5, 1, 4]
[3, 2, 6, 5, 4, 1]
[3, 6, 1, 2, 5, 4]
[3, 6, 1, 5, 2, 4]
[3, 6, 1, 5, 4, 2]
[3, 6, 2, 1, 5, 4]
[3, 6, 2, 5, 1, 4]
[3, 6, 2, 5, 4, 1]
[3, 6, 5, 1, 2, 4]
[3, 6, 5, 1, 4, 2]
[3, 6, 5, 2, 1, 4]
[3, 6, 5, 2, 4, 1]
[3, 6, 5, 4, 1, 2]
[3, 6, 5, 4, 2, 1]
[6, 1, 3, 2, 5, 4]
[6, 1, 3, 5, 2, 4]
[6, 1, 3, 5, 4, 2]
[6, 1, 5, 3, 2, 4]
[6, 1, 5, 3, 4, 2]
[6, 1, 5, 4, 3, 2]
[6, 3, 1, 2, 5, 4]
[6, 3, 1, 5, 2, 4]
[6, 3, 1, 5, 4, 2]
[6, 3, 2, 1, 5, 4]
[6, 3, 2, 5, 1, 4]
[6, 3, 2, 5, 4, 1]
[6, 3, 5, 1, 2, 4]
[6, 3, 5, 1, 4, 2]
[6, 3, 5, 2, 1, 4]
[6, 3, 5, 2, 4, 1]
[6, 3, 5, 4, 1, 2]
[6, 3, 5, 4, 2, 1]
[6, 5, 1, 3, 2, 4]
[6, 5, 1, 3, 4, 2]
[6, 5, 1, 4, 3, 2]
[6, 5, 3, 1, 2, 4]
[6, 5, 3, 1, 4, 2]
[6, 5, 3, 2, 1, 4]
[6, 5, 3, 2, 4, 1]
[6, 5, 3, 4, 1, 2]
[6, 5, 3, 4, 2, 1]
[6, 5, 4, 1, 3, 2]
[6, 5, 4, 3, 1, 2]
[6, 5, 4, 3, 2, 1]

Input number of columns, then column heights in sequence:
3 1 3 3

There are 140 ways to take these 3 columns down.
3×3 Matrix{String}:
"1 "  "2 "  "5 "
"  "  "3 "  "6 "
"  "  "4 "  "7 "

[1, 4, 3, 2, 7, 6, 5]
[1, 4, 3, 7, 2, 6, 5]
[1, 4, 3, 7, 6, 2, 5]
[1, 4, 3, 7, 6, 5, 2]
[1, 4, 7, 3, 2, 6, 5]
[1, 4, 7, 3, 6, 2, 5]
[1, 4, 7, 3, 6, 5, 2]
[1, 4, 7, 6, 3, 2, 5]
[1, 4, 7, 6, 3, 5, 2]
[1, 4, 7, 6, 5, 3, 2]
[1, 7, 4, 3, 2, 6, 5]
[1, 7, 4, 3, 6, 2, 5]
[1, 7, 4, 3, 6, 5, 2]
[1, 7, 4, 6, 3, 2, 5]
[1, 7, 4, 6, 3, 5, 2]
[1, 7, 4, 6, 5, 3, 2]
[1, 7, 6, 4, 3, 2, 5]
[1, 7, 6, 4, 3, 5, 2]
[1, 7, 6, 4, 5, 3, 2]
[1, 7, 6, 5, 4, 3, 2]
[4, 1, 3, 2, 7, 6, 5]
[4, 1, 3, 7, 2, 6, 5]
[4, 1, 3, 7, 6, 2, 5]
[4, 1, 3, 7, 6, 5, 2]
[4, 1, 7, 3, 2, 6, 5]
[4, 1, 7, 3, 6, 2, 5]
[4, 1, 7, 3, 6, 5, 2]
[4, 1, 7, 6, 3, 2, 5]
[4, 1, 7, 6, 3, 5, 2]
[4, 1, 7, 6, 5, 3, 2]
[4, 3, 1, 2, 7, 6, 5]
[4, 3, 1, 7, 2, 6, 5]
[4, 3, 1, 7, 6, 2, 5]
[4, 3, 1, 7, 6, 5, 2]
[4, 3, 2, 1, 7, 6, 5]
[4, 3, 2, 7, 1, 6, 5]
[4, 3, 2, 7, 6, 1, 5]
[4, 3, 2, 7, 6, 5, 1]
[4, 3, 7, 1, 2, 6, 5]
[4, 3, 7, 1, 6, 2, 5]
[4, 3, 7, 1, 6, 5, 2]
[4, 3, 7, 2, 1, 6, 5]
[4, 3, 7, 2, 6, 1, 5]
[4, 3, 7, 2, 6, 5, 1]
[4, 3, 7, 6, 1, 2, 5]
[4, 3, 7, 6, 1, 5, 2]
[4, 3, 7, 6, 2, 1, 5]
[4, 3, 7, 6, 2, 5, 1]
[4, 3, 7, 6, 5, 1, 2]
[4, 3, 7, 6, 5, 2, 1]
[4, 7, 1, 3, 2, 6, 5]
[4, 7, 1, 3, 6, 2, 5]
[4, 7, 1, 3, 6, 5, 2]
[4, 7, 1, 6, 3, 2, 5]
[4, 7, 1, 6, 3, 5, 2]
[4, 7, 1, 6, 5, 3, 2]
[4, 7, 3, 1, 2, 6, 5]
[4, 7, 3, 1, 6, 2, 5]
[4, 7, 3, 1, 6, 5, 2]
[4, 7, 3, 2, 1, 6, 5]
[4, 7, 3, 2, 6, 1, 5]
[4, 7, 3, 2, 6, 5, 1]
[4, 7, 3, 6, 1, 2, 5]
[4, 7, 3, 6, 1, 5, 2]
[4, 7, 3, 6, 2, 1, 5]
[4, 7, 3, 6, 2, 5, 1]
[4, 7, 3, 6, 5, 1, 2]
[4, 7, 3, 6, 5, 2, 1]
[4, 7, 6, 1, 3, 2, 5]
[4, 7, 6, 1, 3, 5, 2]
[4, 7, 6, 1, 5, 3, 2]
[4, 7, 6, 3, 1, 2, 5]
[4, 7, 6, 3, 1, 5, 2]
[4, 7, 6, 3, 2, 1, 5]
[4, 7, 6, 3, 2, 5, 1]
[4, 7, 6, 3, 5, 1, 2]
[4, 7, 6, 3, 5, 2, 1]
[4, 7, 6, 5, 1, 3, 2]
[4, 7, 6, 5, 3, 1, 2]
[4, 7, 6, 5, 3, 2, 1]
[7, 1, 4, 3, 2, 6, 5]
[7, 1, 4, 3, 6, 2, 5]
[7, 1, 4, 3, 6, 5, 2]
[7, 1, 4, 6, 3, 2, 5]
[7, 1, 4, 6, 3, 5, 2]
[7, 1, 4, 6, 5, 3, 2]
[7, 1, 6, 4, 3, 2, 5]
[7, 1, 6, 4, 3, 5, 2]
[7, 1, 6, 4, 5, 3, 2]
[7, 1, 6, 5, 4, 3, 2]
[7, 4, 1, 3, 2, 6, 5]
[7, 4, 1, 3, 6, 2, 5]
[7, 4, 1, 3, 6, 5, 2]
[7, 4, 1, 6, 3, 2, 5]
[7, 4, 1, 6, 3, 5, 2]
[7, 4, 1, 6, 5, 3, 2]
[7, 4, 3, 1, 2, 6, 5]
[7, 4, 3, 1, 6, 2, 5]
[7, 4, 3, 1, 6, 5, 2]
[7, 4, 3, 2, 1, 6, 5]
[7, 4, 3, 2, 6, 1, 5]
[7, 4, 3, 2, 6, 5, 1]
[7, 4, 3, 6, 1, 2, 5]
[7, 4, 3, 6, 1, 5, 2]
[7, 4, 3, 6, 2, 1, 5]
[7, 4, 3, 6, 2, 5, 1]
[7, 4, 3, 6, 5, 1, 2]
[7, 4, 3, 6, 5, 2, 1]
[7, 4, 6, 1, 3, 2, 5]
[7, 4, 6, 1, 3, 5, 2]
[7, 4, 6, 1, 5, 3, 2]
[7, 4, 6, 3, 1, 2, 5]
[7, 4, 6, 3, 1, 5, 2]
[7, 4, 6, 3, 2, 1, 5]
[7, 4, 6, 3, 2, 5, 1]
[7, 4, 6, 3, 5, 1, 2]
[7, 4, 6, 3, 5, 2, 1]
[7, 4, 6, 5, 1, 3, 2]
[7, 4, 6, 5, 3, 1, 2]
[7, 4, 6, 5, 3, 2, 1]
[7, 6, 1, 4, 3, 2, 5]
[7, 6, 1, 4, 3, 5, 2]
[7, 6, 1, 4, 5, 3, 2]
[7, 6, 1, 5, 4, 3, 2]
[7, 6, 4, 1, 3, 2, 5]
[7, 6, 4, 1, 3, 5, 2]
[7, 6, 4, 1, 5, 3, 2]
[7, 6, 4, 3, 1, 2, 5]
[7, 6, 4, 3, 1, 5, 2]
[7, 6, 4, 3, 2, 1, 5]
[7, 6, 4, 3, 2, 5, 1]
[7, 6, 4, 3, 5, 1, 2]
[7, 6, 4, 3, 5, 2, 1]
[7, 6, 4, 5, 1, 3, 2]
[7, 6, 4, 5, 3, 1, 2]
[7, 6, 4, 5, 3, 2, 1]
[7, 6, 5, 1, 4, 3, 2]
[7, 6, 5, 4, 1, 3, 2]
[7, 6, 5, 4, 3, 1, 2]
[7, 6, 5, 4, 3, 2, 1]

Input number of columns, then column heights in sequence:
9 1 2 3 4 5 6 7 8 9

There are 65191584694745586153436251091200000 ways to take these 9 columns down.
```

## Nim

Recursive solution.

The number of elements in the columns are provided as command arguments.

```import std/[os, strutils]

proc sequenceCount(columns: var seq[int]): int =
for icol in 1..columns.high:
if columns[icol] > 0:
dec columns[icol]
inc result, sequenceCount(columns)
inc columns[icol]
if result == 0: result = 1

let ncol = paramCount()
if ncol == 0:
quit "Missing parameters.", QuitFailure
var columns = newSeq[int](ncol + 1)   # We will ignore the first column.
for i in 1..ncol:
let n = paramStr(i).parseInt()
if n < 0:
quit "Wrong number of lanterns.", QuitFailure
columns[i] = n

echo columns.sequenceCount()
```

## Pascal

A console application in Free Pascal, created with the Lazarus IDE.

This solution avoids recursion and calculates the result mathematically. As noted in the Picat solution, the result is a multinomial coefficient, e.g. with columns of length 3, 6, 4 the result is (3 + 6 + 4)!/(3!*6!*4!).

```program LanternProblem;
uses SysUtils;

// Calculate multinomial coefficient, e.g. input array [3, 6, 4]
//   would give (3 + 6 + 4)! / (3!*6!*4!).
// Result is calculated as a product of binomial coefficients.
function Multinomial( a : array of integer) : UInt64;
var
n, i, j, k : integer;
b : array of integer;   // sorted copy of ionput
row : array of integer; // start of row in Pascal's triangle
begin
result := 1; // in case of trivial input
n := Length( a);
if (n <= 1) then exit;

// Copy caller's array to local array in descending order
SetLength( b, n);
for j := 0 to n - 1 do begin
k := j;
while (k > 0) and (b[k - 1] < a[j]) do begin
b[k] := b[k - 1];  dec(k);
end;
b[k] := a[j];
end;

// Zero entries don't affect the result, so remove them
while (n > 0) and (b[n - 1] = 0) do dec(n);
if (n <= 1) then exit;

// Non-trivial input, do the calculation by means of Pascal's triangle
SetLength( row, b[1] + 1);
row[0] := 1;
for k := 1 to b[1] do row[k] := 0;
for i := 1 to b[0] + b[1] do begin
for k := b[1] downto 1 do inc( row[k], row[k - 1]);
end;
result := row[b[1]];  // first binomial coefficient

// Since b[1] >= b[2] >= b[3] ... there are always enough valid terms
//   in the row to allow calculation of the next binomial coefficient.
for j := 2 to n - 1 do begin
for i := 1 to b[j] do begin
for k := b[1] downto 1 do inc( row[k], row[k - 1]);
end;
result := result*row[b[j]]; // multiply by next binomial coefficient
end;
end;

// Prompt user for non-negative integer.
// Avoid raising exception when user input isn't an integer.
function UserInt( const prompt : string) : integer;
var
userInput : string;
inputOK : boolean;
begin
repeat
Write( prompt, ' ');
inputOK := SysUtils.TryStrToInt( userInput, result) and (result >= 0);
if not inputOK then WriteLn( 'Try again');
until inputOK;
end;

// Main routine
var
nrCols, j : integer;
counts : array of integer;
begin
repeat
nrCols := UserInt( 'Number of columns (0 to quit)?');
if nrCols = 0 then exit;
SetLength( counts, nrCols);
for j := 0 to nrCols - 1 do
counts[j] := UserInt( SysUtils.Format('How many in column %d?',
[j + 1])); // column numbers 1-based for user
Write( 'Columns are ');
for j := 0 to nrCols - 1 do Write(' ', counts[j]);
WriteLn( ',  number of ways = ', Multinomial(counts));
until false;
end.
```
Output:
```Number of columns (0 to quit)? 3
How many in column 1? 1
How many in column 2? 2
How many in column 3? 3
Columns are  1 2 3,  number of ways = 60
[input omitted from now on]
Columns are  1 2 3 4,  number of ways = 12600
Columns are  1 2 3 4 5,  number of ways = 37837800
Columns are  1 2 3 4 5 6,  number of ways = 2053230379200
Columns are  1 2 3 4 5 6 7,  number of ways = 2431106898187968000
Columns are  1 3 3,  number of ways = 140
```

## Perl

```#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Solve_hanging_lantern_problem
use warnings;

\$_ = 'a bc def';

find(\$_, '');

sub find
{
my (\$in, \$found) = @_;
find( \$` . \$', \$found . \$& ) while \$in =~ /\w\b/g;
\$in =~ /\w/ or \$answer .= '[' . \$found =~ s/\B/,/gr . "]\n";
}
```
Output:
```count = 60
[a,c,b,f,e,d]
[a,c,f,b,e,d]
[a,c,f,e,b,d]
[a,c,f,e,d,b]
[a,f,c,b,e,d]
[a,f,c,e,b,d]
[a,f,c,e,d,b]
[a,f,e,c,b,d]
[a,f,e,c,d,b]
[a,f,e,d,c,b]
[c,a,b,f,e,d]
[c,a,f,b,e,d]
[c,a,f,e,b,d]
[c,a,f,e,d,b]
[c,b,a,f,e,d]
[c,b,f,a,e,d]
[c,b,f,e,a,d]
[c,b,f,e,d,a]
[c,f,a,b,e,d]
[c,f,a,e,b,d]
[c,f,a,e,d,b]
[c,f,b,a,e,d]
[c,f,b,e,a,d]
[c,f,b,e,d,a]
[c,f,e,a,b,d]
[c,f,e,a,d,b]
[c,f,e,b,a,d]
[c,f,e,b,d,a]
[c,f,e,d,a,b]
[c,f,e,d,b,a]
[f,a,c,b,e,d]
[f,a,c,e,b,d]
[f,a,c,e,d,b]
[f,a,e,c,b,d]
[f,a,e,c,d,b]
[f,a,e,d,c,b]
[f,c,a,b,e,d]
[f,c,a,e,b,d]
[f,c,a,e,d,b]
[f,c,b,a,e,d]
[f,c,b,e,a,d]
[f,c,b,e,d,a]
[f,c,e,a,b,d]
[f,c,e,a,d,b]
[f,c,e,b,a,d]
[f,c,e,b,d,a]
[f,c,e,d,a,b]
[f,c,e,d,b,a]
[f,e,a,c,b,d]
[f,e,a,c,d,b]
[f,e,a,d,c,b]
[f,e,c,a,b,d]
[f,e,c,a,d,b]
[f,e,c,b,a,d]
[f,e,c,b,d,a]
[f,e,c,d,a,b]
[f,e,c,d,b,a]
[f,e,d,a,c,b]
[f,e,d,c,a,b]
[f,e,d,c,b,a]
```

## Phix

```with javascript_semantics
include mpfr.e
function get_lantern(mpz z, sequence s, bool bJustCount=true, integer na=-1)
if bJustCount then
sequence l = apply(s,length)
mpz_fac_ui(z,sum(l))
mpz f = mpz_init()
for d in l do
mpz_fac_ui(f,d)
mpz_fdiv_q(z,z,f)
end for
return 0
end if
if na=-1 then na = sum(apply(s,length)) end if
if na=0 then
mpz_set_si(z,1)
return {""}
end if
s = deep_copy(s)
sequence res = {}
for i=1 to length(s) do
if length(s[i]) then
integer si = s[i][\$]
s[i] = s[i][1..\$-1]
mpz z2 = mpz_init()
object r = get_lantern(z2, s, false, na-1)
for k=1 to length(r) do
res = append(res,si&r[k])
end for
s[i] &= si
end if
end for
return res
end function

procedure test(sequence s, bool bJustCount=true)
mpz z = mpz_init()
object r = get_lantern(z,s,bJustCount)
string sj = join(s,", "),
sz = mpz_get_str(z)
if bJustCount then
printf(1,"%s = %s\n",{sj,sz})
else
string rj = join_by(r,1,10,",")
printf(1,"%s = %s:\n%s\n",{sj,sz,rj})
end if
end procedure

test({"1"},false)
test({"1","23"},false)
test({"1","23","456"},false)
test({"1","234","567"})
test({"1234567890","ABCDEFGHIJKLMN","OPQRSTUVWXYZ"})
sequence s = {"1",
"23",
"456",
"7890",
"ABCDE",
"FGHIJK",
"LMNOPQR",
"STUVWXYZ",
"abcdefghi"}
for i=1 to 9 do
test(s[1..i])
end for
```
Output:
```1 = 1:
1

1, 23 = 3:
132,312,321

1, 23, 456 = 60:
132654,136254,136524,136542,163254,163524,163542,165324,165342,165432
312654,316254,316524,316542,321654,326154,326514,326541,361254,361524
361542,362154,362514,362541,365124,365142,365214,365241,365412,365421
613254,613524,613542,615324,615342,615432,631254,631524,631542,632154
632514,632541,635124,635142,635214,635241,635412,635421,651324,651342
651432,653124,653142,653214,653241,653412,653421,654132,654312,654321

1, 234, 567 = 140
1234567890, ABCDEFGHIJKLMN, OPQRSTUVWXYZ = 2454860399191200
1 = 1
1, 23 = 3
1, 23, 456 = 60
1, 23, 456, 7890 = 12600
1, 23, 456, 7890, ABCDE = 37837800
1, 23, 456, 7890, ABCDE, FGHIJK = 2053230379200
1, 23, 456, 7890, ABCDE, FGHIJK, LMNOPQR = 2431106898187968000
1, 23, 456, 7890, ABCDE, FGHIJK, LMNOPQR, STUVWXYZ = 73566121315513295589120000
1, 23, 456, 7890, ABCDE, FGHIJK, LMNOPQR, STUVWXYZ, abcdefghi = 65191584694745586153436251091200000
```

## Picat

Translation of: Python
```main =>
run_lantern().

run_lantern() =>
A = [],
foreach(_ in 1..N)
end,
println(A),
println(lantern(A)),
nl.

table
lantern(A) = Res =>
Arr = copy_term(A),
Res = 0,
foreach(I in 1..Arr.len)
if Arr[I] != 0 then
Arr[I] := Arr[I] - 1,
Res := Res + lantern(Arr),
Arr[I] := Arr[I] + 1
end
end,
if Res == 0 then
Res := 1
end.```

Some tests:

```main =>
A = [1,2,3],
println(lantern(A)),
foreach(N in 1..8)
println(1..N=lantern(1..N))
end,
nl.```
Output:
```60
[1] = 1
[1,2] = 3
[1,2,3] = 60
[1,2,3,4] = 12600
[1,2,3,4,5] = 37837800
[1,2,3,4,5,6] = 2053230379200
[1,2,3,4,5,6,7] = 2431106898187968000
[1,2,3,4,5,6,7,8] = 73566121315513295589120000```

The sequence of lantern(1..N) is the OEIS sequence A022915 ("Multinomial coefficients (0, 1, ..., n)! = C(n+1,2)!/(0!*1!*2!*...*n!)").

## Python

### Recursive version

Translation of: Visual Basic
```def getLantern(arr):
res = 0
for i in range(0, n):
if arr[i] != 0:
arr[i] -= 1
res += getLantern(arr)
arr[i] += 1
if res == 0:
res = 1
return res

a = []
n = int(input())
for i in range(0, n):
a.append(int(input()))
print(getLantern(a))
```

### Math solution

```import math
n = int(input())
a = []
tot = 0
for i in range(0, n):
a.append(int(input()))
tot += a[i]
res = math.factorial(tot)
for i in range(0, n):
res /= math.factorial(a[i])
print(int(res))
```

### Showing Sequences

```def seq(x):
if not any(x):
yield tuple()

for i, v in enumerate(x):
if v:
for s in seq(x[:i] + [v - 1] + x[i+1:]):
yield (i+1,) + s

# an example
for x in seq([1, 2, 3]):
print(x)
```

## Quackery

```  [ 1 swap times [ i 1+ * ] ]   is !     ( n --> n )

[ 0 over witheach + !
1 rot witheach [ ! * ]
/ ]                         is solve ( [ --> n )

[ []
\$ "Number of columns? "
input quackery times
[ \$ "How many lanterns? "
input quackery join ]
solve
say "Removal sequences: "
echo ]                      is task   (  -->   )```
Output:
```Number of columns? 8
How many lanterns? 1
How many lanterns? 2
How many lanterns? 3
How many lanterns? 4
How many lanterns? 5
How many lanterns? 6
How many lanterns? 7
How many lanterns? 8
Removal sequences: 73566121315513295589120000
```

## Raku

Note: All of these solutions accept the list of column sizes as command-line arguments and infer the number of columns from the number of sizes provided, rather than requiring that a count be supplied as an extra distinct parameter.

### Directly computing the count

Translation of: Pascal

If all we need is the count, then we can compute that directly:

```unit sub MAIN(*@columns);

sub postfix:<!>(\$n) { [*] 1..\$n }

say [+](@columns)! / [*](@columns»!);
```
Output:
```\$ raku lanterns.raku 1 2 3
60```

### Sequence as column numbers

Translation of: Julia

If we want to list all of the sequences, we have to do some more work. This version outputs the sequences as lists of column numbers (assigned from 1 to N left to right); at each step the bottommost lantern from the numbered column is removed.

```unit sub MAIN(*@columns, :v(:\$verbose)=False);

my @sequences = @columns
. pairs
. map({ (.key+1) xx .value })
. flat
. permutations
. map( *.join(',') )
. unique;

if (\$verbose) {
say "There are {+@sequences} possible takedown sequences:";
say "[\$_]" for @sequences;
} else {
say +@sequences;
}
```
Output:
```\$ raku lanterns.raku --verbose 1 2 3
There are 60 possible takedown sequences:
[1,2,2,3,3,3]
[1,2,3,2,3,3]
[1,2,3,3,2,3]
[1,2,3,3,3,2]
[1,3,2,2,3,3]
[1,3,2,3,2,3]
...
[3,3,2,2,3,1]
[3,3,2,3,1,2]
[3,3,2,3,2,1]
[3,3,3,1,2,2]
[3,3,3,2,1,2]
[3,3,3,2,2,1]```

### Sequence as lantern numbers

If we want individually-numbered lanterns in the sequence instead of column numbers, as in the example given in the task description, that requires yet more work:

```unit sub MAIN(*@columns, :v(:\$verbose)=False);

my @sequences = @columns
. pairs
. map({ (.key+1) xx .value })
. flat
. permutations
. map( *.join(',') )
. unique;

if (\$verbose) {
my @offsets = |0,|(1..@columns).map: { [+] @columns[0..\$_-1] };
my @matrix;
for ^@columns.max -> \$i {
for ^@columns -> \$j {
my \$value = \$i < @columns[\$j] ?? (\$i+@offsets[\$j]+1) !! Nil;
@matrix[\$j][\$i] = \$value if \$value;;
print "\t" ~ (\$value // " ");
}
say '';
}
say "There are {+@sequences} possible takedown sequences:";
for @sequences».split(',') -> @seq {
my @work = @matrix».clone;
my \$seq = '[';
for @seq -> \$col {
\$seq ~= @work[\$col-1].pop ~ ',';
}
\$seq ~~ s/','\$/]/;
say \$seq;
}
} else {
say +@sequences;
}
```
Output:
```\$ raku lanterns.raku -v 1 2 3
1       2       4
3       5
6
There are 60 possible takedown sequences:
[1,3,2,6,5,4]
[1,3,6,2,5,4]
[1,3,6,5,2,4]
...
[6,5,4,1,3,2]
[6,5,4,3,1,2]
[6,5,4,3,2,1]```

## Ruby

### Directly computing the count

Compute the count directly:

```Factorial = Hash.new{|h, k| h[k] = k * h[k-1] } # a memoized factorial
Factorial[0] = 1

def count_perms_with_reps(ar)
Factorial[ar.sum] / ar.inject{|prod, m| prod * Factorial[m]}
end

ar, input = [], ""
puts "Input column heights in sequence (empty line to end input):"
ar << input.to_i until (input=gets) == "\n"
puts "There are #{count_perms_with_reps(ar)} ways to take these #{ar.size} columns down."
```
Output:
```Input column heights in sequence (empty line to end input):
1
2
3
4
5
6
7
8

There are 73566121315513295589120000 ways to take these 8 columns down.
```

## Uiua

Works with: Uiua version 0.10.0
```Fac ← /×+1⇡
Lant ← ÷⊃(/(×⊙Fac)|Fac/+)

Lant [1 2 3]
Lant [1 3 3]
Lant [1 3 3 5 7]```
Output:
```60
140
5587021440
```

## Wren

### Version 1

Translation of: Python

The result for n == 5 is slow to emerge.

```var lantern // recursive function
lantern = Fn.new { |n, a|
var count = 0
for (i in 0...n) {
if (a[i] != 0) {
a[i] = a[i] - 1
count = count + lantern.call(n, a)
a[i] = a[i] + 1
}
}
if (count == 0) count = 1
return count
}

System.print("Number of permutations for n (<= 5) groups and lanterns per group [1..n]:")
var n = 0
for (i in 1..5) {
var a = (1..i).toList
n = n + 1
System.print("%(a) => %(lantern.call(n, a))")
}
```
Output:
```Number of permutations for n (<= 5) groups and lanterns per group [1..n]:
[1] => 1
[1, 2] => 3
[1, 2, 3] => 60
[1, 2, 3, 4] => 12600
[1, 2, 3, 4, 5] => 37837800
```

### Version 2

Library: Wren-perm
Library: Wren-big

Alternatively, using library methods.

```import "./perm" for Perm
import "./big" for BigInt

var listPerms = Fn.new { |a, rowSize|
var lows = List.filled(a.count, 0)
var sum = 0
var mlist = []
for (i in 0...a.count) {
sum = sum + a[i]
lows[i] = sum
mlist = mlist + [i+1] * a[i]
}
var n = Perm.countDistinct(sum, a)
System.print("\nList of %(n) permutations for %(a.count) groups and lanterns per group %(a):")
var count = 0
for (p in Perm.listDistinct(mlist)) {
var curr = lows.toList
var perm = List.filled(sum, 0)
for (i in 0...sum) {
perm[i] = curr[p[i]-1]
curr[p[i]-1] = curr[p[i]-1] - 1
}
System.write("%(perm) ")
count = count + 1
if (count % rowSize == 0) System.print()
}
if (count % rowSize != 0) System.print()
}

System.print("Number of permutations for the lanterns per group shown:")
var n = 0
for (i in 1..9) {
var a = (1..i).toList
n = n + i
System.print("%(a) => %(BigInt.multinomial(n, a))")
}
var a = [1, 3, 3]
System.print("%(a) => %(BigInt.multinomial(7, a))")
a = [10, 14, 12]
System.print("%(a) => %(BigInt.multinomial(36, a))")
listPerms.call([1, 2, 3], 4)
listPerms.call([1, 3, 3], 3)
```
Output:
```Number of permutations for the lanterns per group shown:
[1] => 1
[1, 2] => 3
[1, 2, 3] => 60
[1, 2, 3, 4] => 12600
[1, 2, 3, 4, 5] => 37837800
[1, 2, 3, 4, 5, 6] => 2053230379200
[1, 2, 3, 4, 5, 6, 7] => 2431106898187968000
[1, 2, 3, 4, 5, 6, 7, 8] => 73566121315513295589120000
[1, 2, 3, 4, 5, 6, 7, 8, 9] => 65191584694745586153436251091200000
[1, 3, 3] => 140
[10, 14, 12] => 2454860399191200

List of 60 permutations for 3 groups and lanterns per group [1, 2, 3]:
[1, 3, 2, 6, 5, 4] [1, 3, 6, 2, 5, 4] [1, 3, 6, 5, 2, 4] [1, 3, 6, 5, 4, 2] [1, 6, 3, 2, 5, 4]
[1, 6, 3, 5, 2, 4] [1, 6, 3, 5, 4, 2] [1, 6, 5, 3, 2, 4] [1, 6, 5, 3, 4, 2] [1, 6, 5, 4, 3, 2]
[3, 1, 2, 6, 5, 4] [3, 1, 6, 2, 5, 4] [3, 1, 6, 5, 2, 4] [3, 1, 6, 5, 4, 2] [3, 2, 1, 6, 5, 4]
[3, 2, 6, 1, 5, 4] [3, 2, 6, 5, 1, 4] [3, 2, 6, 5, 4, 1] [3, 6, 2, 1, 5, 4] [3, 6, 2, 5, 1, 4]
[3, 6, 2, 5, 4, 1] [3, 6, 1, 2, 5, 4] [3, 6, 1, 5, 2, 4] [3, 6, 1, 5, 4, 2] [3, 6, 5, 1, 2, 4]
[3, 6, 5, 1, 4, 2] [3, 6, 5, 2, 1, 4] [3, 6, 5, 2, 4, 1] [3, 6, 5, 4, 2, 1] [3, 6, 5, 4, 1, 2]
[6, 3, 2, 1, 5, 4] [6, 3, 2, 5, 1, 4] [6, 3, 2, 5, 4, 1] [6, 3, 1, 2, 5, 4] [6, 3, 1, 5, 2, 4]
[6, 3, 1, 5, 4, 2] [6, 3, 5, 1, 2, 4] [6, 3, 5, 1, 4, 2] [6, 3, 5, 2, 1, 4] [6, 3, 5, 2, 4, 1]
[6, 3, 5, 4, 2, 1] [6, 3, 5, 4, 1, 2] [6, 1, 3, 2, 5, 4] [6, 1, 3, 5, 2, 4] [6, 1, 3, 5, 4, 2]
[6, 1, 5, 3, 2, 4] [6, 1, 5, 3, 4, 2] [6, 1, 5, 4, 3, 2] [6, 5, 3, 1, 2, 4] [6, 5, 3, 1, 4, 2]
[6, 5, 3, 2, 1, 4] [6, 5, 3, 2, 4, 1] [6, 5, 3, 4, 2, 1] [6, 5, 3, 4, 1, 2] [6, 5, 1, 3, 2, 4]
[6, 5, 1, 3, 4, 2] [6, 5, 1, 4, 3, 2] [6, 5, 4, 1, 3, 2] [6, 5, 4, 3, 1, 2] [6, 5, 4, 3, 2, 1]

List of 140 permutations for 3 groups and lanterns per group [1, 3, 3]:
[1, 4, 3, 2, 7, 6, 5] [1, 4, 3, 7, 2, 6, 5] [1, 4, 3, 7, 6, 2, 5] [1, 4, 3, 7, 6, 5, 2]
[1, 4, 7, 3, 2, 6, 5] [1, 4, 7, 3, 6, 2, 5] [1, 4, 7, 3, 6, 5, 2] [1, 4, 7, 6, 3, 2, 5]
[1, 4, 7, 6, 3, 5, 2] [1, 4, 7, 6, 5, 3, 2] [1, 7, 4, 3, 2, 6, 5] [1, 7, 4, 3, 6, 2, 5]
[1, 7, 4, 3, 6, 5, 2] [1, 7, 4, 6, 3, 2, 5] [1, 7, 4, 6, 3, 5, 2] [1, 7, 4, 6, 5, 3, 2]
[1, 7, 6, 4, 3, 2, 5] [1, 7, 6, 4, 3, 5, 2] [1, 7, 6, 4, 5, 3, 2] [1, 7, 6, 5, 4, 3, 2]
[4, 1, 3, 2, 7, 6, 5] [4, 1, 3, 7, 2, 6, 5] [4, 1, 3, 7, 6, 2, 5] [4, 1, 3, 7, 6, 5, 2]
[4, 1, 7, 3, 2, 6, 5] [4, 1, 7, 3, 6, 2, 5] [4, 1, 7, 3, 6, 5, 2] [4, 1, 7, 6, 3, 2, 5]
[4, 1, 7, 6, 3, 5, 2] [4, 1, 7, 6, 5, 3, 2] [4, 3, 1, 2, 7, 6, 5] [4, 3, 1, 7, 2, 6, 5]
[4, 3, 1, 7, 6, 2, 5] [4, 3, 1, 7, 6, 5, 2] [4, 3, 2, 1, 7, 6, 5] [4, 3, 2, 7, 1, 6, 5]
[4, 3, 2, 7, 6, 1, 5] [4, 3, 2, 7, 6, 5, 1] [4, 3, 7, 2, 1, 6, 5] [4, 3, 7, 2, 6, 1, 5]
[4, 3, 7, 2, 6, 5, 1] [4, 3, 7, 1, 2, 6, 5] [4, 3, 7, 1, 6, 2, 5] [4, 3, 7, 1, 6, 5, 2]
[4, 3, 7, 6, 1, 2, 5] [4, 3, 7, 6, 1, 5, 2] [4, 3, 7, 6, 2, 1, 5] [4, 3, 7, 6, 2, 5, 1]
[4, 3, 7, 6, 5, 2, 1] [4, 3, 7, 6, 5, 1, 2] [4, 7, 3, 2, 1, 6, 5] [4, 7, 3, 2, 6, 1, 5]
[4, 7, 3, 2, 6, 5, 1] [4, 7, 3, 1, 2, 6, 5] [4, 7, 3, 1, 6, 2, 5] [4, 7, 3, 1, 6, 5, 2]
[4, 7, 3, 6, 1, 2, 5] [4, 7, 3, 6, 1, 5, 2] [4, 7, 3, 6, 2, 1, 5] [4, 7, 3, 6, 2, 5, 1]
[4, 7, 3, 6, 5, 2, 1] [4, 7, 3, 6, 5, 1, 2] [4, 7, 1, 3, 2, 6, 5] [4, 7, 1, 3, 6, 2, 5]
[4, 7, 1, 3, 6, 5, 2] [4, 7, 1, 6, 3, 2, 5] [4, 7, 1, 6, 3, 5, 2] [4, 7, 1, 6, 5, 3, 2]
[4, 7, 6, 3, 1, 2, 5] [4, 7, 6, 3, 1, 5, 2] [4, 7, 6, 3, 2, 1, 5] [4, 7, 6, 3, 2, 5, 1]
[4, 7, 6, 3, 5, 2, 1] [4, 7, 6, 3, 5, 1, 2] [4, 7, 6, 1, 3, 2, 5] [4, 7, 6, 1, 3, 5, 2]
[4, 7, 6, 1, 5, 3, 2] [4, 7, 6, 5, 1, 3, 2] [4, 7, 6, 5, 3, 1, 2] [4, 7, 6, 5, 3, 2, 1]
[7, 4, 3, 2, 1, 6, 5] [7, 4, 3, 2, 6, 1, 5] [7, 4, 3, 2, 6, 5, 1] [7, 4, 3, 1, 2, 6, 5]
[7, 4, 3, 1, 6, 2, 5] [7, 4, 3, 1, 6, 5, 2] [7, 4, 3, 6, 1, 2, 5] [7, 4, 3, 6, 1, 5, 2]
[7, 4, 3, 6, 2, 1, 5] [7, 4, 3, 6, 2, 5, 1] [7, 4, 3, 6, 5, 2, 1] [7, 4, 3, 6, 5, 1, 2]
[7, 4, 1, 3, 2, 6, 5] [7, 4, 1, 3, 6, 2, 5] [7, 4, 1, 3, 6, 5, 2] [7, 4, 1, 6, 3, 2, 5]
[7, 4, 1, 6, 3, 5, 2] [7, 4, 1, 6, 5, 3, 2] [7, 4, 6, 3, 1, 2, 5] [7, 4, 6, 3, 1, 5, 2]
[7, 4, 6, 3, 2, 1, 5] [7, 4, 6, 3, 2, 5, 1] [7, 4, 6, 3, 5, 2, 1] [7, 4, 6, 3, 5, 1, 2]
[7, 4, 6, 1, 3, 2, 5] [7, 4, 6, 1, 3, 5, 2] [7, 4, 6, 1, 5, 3, 2] [7, 4, 6, 5, 1, 3, 2]
[7, 4, 6, 5, 3, 1, 2] [7, 4, 6, 5, 3, 2, 1] [7, 1, 4, 3, 2, 6, 5] [7, 1, 4, 3, 6, 2, 5]
[7, 1, 4, 3, 6, 5, 2] [7, 1, 4, 6, 3, 2, 5] [7, 1, 4, 6, 3, 5, 2] [7, 1, 4, 6, 5, 3, 2]
[7, 1, 6, 4, 3, 2, 5] [7, 1, 6, 4, 3, 5, 2] [7, 1, 6, 4, 5, 3, 2] [7, 1, 6, 5, 4, 3, 2]
[7, 6, 4, 3, 1, 2, 5] [7, 6, 4, 3, 1, 5, 2] [7, 6, 4, 3, 2, 1, 5] [7, 6, 4, 3, 2, 5, 1]
[7, 6, 4, 3, 5, 2, 1] [7, 6, 4, 3, 5, 1, 2] [7, 6, 4, 1, 3, 2, 5] [7, 6, 4, 1, 3, 5, 2]
[7, 6, 4, 1, 5, 3, 2] [7, 6, 4, 5, 1, 3, 2] [7, 6, 4, 5, 3, 1, 2] [7, 6, 4, 5, 3, 2, 1]
[7, 6, 1, 4, 3, 2, 5] [7, 6, 1, 4, 3, 5, 2] [7, 6, 1, 4, 5, 3, 2] [7, 6, 1, 5, 4, 3, 2]
[7, 6, 5, 4, 1, 3, 2] [7, 6, 5, 4, 3, 1, 2] [7, 6, 5, 4, 3, 2, 1] [7, 6, 5, 1, 4, 3, 2]
```

## XPL0

```char N, Column, Sequences, I, Lanterns;

proc Tally(Level);
char Level, Col;
[for Col:= 0 to N-1 do
if Column(Col) > 0 then
[Column(Col):= Column(Col)-1;
if Level = Lanterns-1 then Sequences:= Sequences+1
else Tally(Level+1);
Column(Col):= Column(Col)+1;
];
];

[Sequences:= 0;  Lanterns:= 0;
N:= IntIn(0);
Column:= Reserve(N);
for I:= 0 to N-1 do
[Column(I):= IntIn(0);
Lanterns:= Lanterns + Column(I);
];
Tally(0);
IntOut(0, Sequences);
]```
Output:
```5
1 3 5 2 4
37837800
```