Making picture albums (piling in bins 4/4)

Mike has 12 pictures and he wants to put them in some albums, which are identical when empty.

He has to decide how many albums he wants, from 1 album to 11 albums (he just doesn't want to have 12 albums with one picture). For each album, he has to group some picture and decide in which order to put them in the album.

How many ways does have Mike to create some albums with his 12 pictures?


Part of the Piling distinct objects in bins set.


The answer is 12470162232.

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1 solution

Laurent Shorts
Feb 5, 2017

For each number of album i i , Mike has 12 i \left\lfloor \begin{array}{c}12\\i\end{array}\right\rfloor ways to make them. (This is a Lah number , see answer to problem 3/4 .)

We then sum for each possible number of album: i = 1 11 12 i = 1 2 47 0 16 2 232 \sum^{11}_{i=1}\left\lfloor \begin{array}{c}12\\i\end{array}\right\rfloor=12'470'162'232 . (Detail on comment below.)

Number of albums Number of ways
1 12 1 = 47 9 00 1 600 ( = 12 ! ) \left\lfloor \begin{array}{c}12\\1\end{array}\right\rfloor=479'001'600 (=12!)
2 12 2 = 2 63 4 50 8 800 \left\lfloor \begin{array}{c}12\\2\end{array}\right\rfloor=2'634'508'800
3 12 3 = 4 39 0 84 8 000 \left\lfloor \begin{array}{c}12\\3\end{array}\right\rfloor=4'390'848'000
4 12 4 = 3 29 3 13 6 000 \left\lfloor \begin{array}{c}12\\4\end{array}\right\rfloor=3'293'136'000
5 12 5 = 1 31 7 25 4 400 \left\lfloor \begin{array}{c}12\\5\end{array}\right\rfloor=1'317'254'400
6 12 6 = 30 7 35 9 360 \left\lfloor \begin{array}{c}12\\6\end{array}\right\rfloor=307'359'360
7 12 7 = 4 3 90 8 480 \left\lfloor \begin{array}{c}12\\7\end{array}\right\rfloor=43'908'480
8 12 8 = 3 92 0 400 \left\lfloor \begin{array}{c}12\\8\end{array}\right\rfloor=3'920'400
9 12 9 = 21 7 800 \left\lfloor \begin{array}{c}12\\9\end{array}\right\rfloor=217'800
10 12 10 = 7 260 \left\lfloor \begin{array}{c}12\\10\end{array}\right\rfloor=7'260
11 12 11 = 132 \left\lfloor \begin{array}{c}12\\11\end{array}\right\rfloor=132
12 12 12 = 1 \left\lfloor \begin{array}{c}12\\12\end{array}\right\rfloor=1 , not counted

Laurent Shorts - 4 years, 4 months ago

Nice solution . Well , is there any closed form of this summation ? @Laurent Shorts

Ujjwal Mani Tripathi - 4 years, 3 months ago

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Thank you. There's no closed form to my knowledge, but I can be mistaken.

Laurent Shorts - 4 years, 3 months ago

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