# There are six different books distributed to 3 boys. Each boy at least got one book. How many ways of the combinations?

1920

#### Explanation:

There are a few cases we need to explore:

• If each boy gets only 1 book each (3 books total), we can set up a permutation (order matters):

P_(n,k)=(n!)/((n-k)!); n="population", k="picks"

P_(6,3)=(6!)/(3!)=6xx5xx4=120

• If each boy gets 2 books each (6 books total), we can set up a permutation:

P_(6,6)=(6!)/(0!)=720

• If one boy gets 2 books and the others get 1 each (4 books total), we can set up a permutation expression that starts with each boy getting one book, and then the remaining 4th book can go to any one of the three boys:

P_(6,3)xxP_(3,1)=(6!)/(3!)xx(3!)/(2!)=6xx5xx4xx3=360

• If two boys gets 2 books and the other gets 1 (5 books total), we can set up a permutation expression that starts with each boy getting one book, and then the remaining 4th and 5th books can go to any two of the three boys:

P_(6,3)xxP_(3,2)=(6!)/(3!)xx(3!)/(1!)=6xx5xx4xx3xx2=720

And now we add up the cases:

$120 + 720 + 360 + 720 = 1920$