Question

In how many different ways can a man divide 7 different gifts among his 3 children if the eldest is to receive 3 gifts and the others 2 each

Answer 1

The order in which gifts are received doesn't matter - if child X gets toy 1 then toy 2, it's the same as giving child X toy 2 then toy 1 - so we are counting combinations.

The eldest child receives 3 of the 7 gifts, so they have

$\dbinom 73 = \dfrac{7!}{3!(7-3)!} = 35$

possible choices of gifts.

The next child receives 2 of the remaining 4 gifts, so they have

$\dbinom 42 = \dfrac{4!}{2!(4-2)!} = 6$

choices.

The last child receives the remaining 2 gifts, and there is only

$\dbinom22 = \dfrac{2!}{2!(2-2)!} = 1$

way to select the gifts for them.

By the multiplication using, the total number of ways of distributing 7 gifts among 3 children in the prescribed way is 35 • 6 • 1 = 210.