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?
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Number of albums | Number of ways |
1 | ⌊ 1 2 1 ⌋ = 4 7 9 ′ 0 0 1 ′ 6 0 0 ( = 1 2 ! ) |
2 | ⌊ 1 2 2 ⌋ = 2 ′ 6 3 4 ′ 5 0 8 ′ 8 0 0 |
3 | ⌊ 1 2 3 ⌋ = 4 ′ 3 9 0 ′ 8 4 8 ′ 0 0 0 |
4 | ⌊ 1 2 4 ⌋ = 3 ′ 2 9 3 ′ 1 3 6 ′ 0 0 0 |
5 | ⌊ 1 2 5 ⌋ = 1 ′ 3 1 7 ′ 2 5 4 ′ 4 0 0 |
6 | ⌊ 1 2 6 ⌋ = 3 0 7 ′ 3 5 9 ′ 3 6 0 |
7 | ⌊ 1 2 7 ⌋ = 4 3 ′ 9 0 8 ′ 4 8 0 |
8 | ⌊ 1 2 8 ⌋ = 3 ′ 9 2 0 ′ 4 0 0 |
9 | ⌊ 1 2 9 ⌋ = 2 1 7 ′ 8 0 0 |
10 | ⌊ 1 2 1 0 ⌋ = 7 ′ 2 6 0 |
11 | ⌊ 1 2 1 1 ⌋ = 1 3 2 |
12 | ⌊ 1 2 1 2 ⌋ = 1 , not counted |
Nice solution . Well , is there any closed form of this summation ? @Laurent Shorts
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Thank you. There's no closed form to my knowledge, but I can be mistaken.
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For each number of album i , Mike has ⌊ 1 2 i ⌋ 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 1 1 ⌊ 1 2 i ⌋ = 1 2 ′ 4 7 0 ′ 1 6 2 ′ 2 3 2 . (Detail on comment below.)