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

1 Answer

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):

Pn,k=n!(nk)!;n=population,k=picks

P6,3=6!3!=6×5×4=120

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

P6,6=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:

P6,3×P3,1=6!3!×3!2!=6×5×4×3=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:

P6,3×P3,2=6!3!×3!1!=6×5×4×3×2=720

And now we add up the cases:

120+720+360+720=1920