In how many ways can $3$ red, $2$ blue, and $1$ yellow gift be distributed to eight people if no person receives more than one gift?

Question

We have $6$ gifts for $8$ people: $3$ gifts are red, $2$ are blue, and $1$ is yellow. In how many ways can we distribute the gifts? (one person gets at most one gift; assume gifts of the same color are the same)

I'm a little confused - haven't seen a problem like this one before in the course that I'm in... I'm assuming it's related to the partition formula/permutations with indistinct objects $$\frac{8!}{3! 2! 1!}$$ but should $\binom{8}{6}$ (permutation) be in the numerator?

Answer 1

There are $\binom{8}{3}$ ways to choose the recipients of the red gifts. There are $\binom{5}{2}$ ways to choose the recipients of the blue gifts from the remaining five people. There are $\binom{3}{1}$ ways to choose the recipient of the yellow gift from the remaining three people. Hence, the number of ways to distribute the gifts is $$\binom{8}{3}\binom{5}{2}\binom{3}{1} = \frac{8!}{3!5!} \cdot \frac{5!}{2!3!} \cdot \frac{3!}{1!2!} = \frac{8!}{3!2!1!2!}$$ As @cgiovanardi indicated in the comments, the second factor of $2!$ in the denominator is attributable to the two people who each receive no gifts.

Answer 2

Regarding your confusion, and your remarks in the question, you may find it easier to think of it as choosing those to whom gifts will be given, and then distributing the gifts to the chosen ones, thus

$\dbinom86 \times \dfrac{6!}{3!2!1!}$

Answer 3

Introduce two non-gifts, so the number of items to be distributed equals the number of people. Then the number of ways to distribute the items is the number of permutations of $8$ objects where there are indistinguishable objects, with counts $\{3,2,1,2\}$, which is $$\frac{8!}{3! 2! 1! 2!}$$