Here for (a), I think the solution would be 4! × C(15,4) = 32760. Is that correct? As for (b), I am confused. Should I use C(n+r-1,r) here? Or is it something else? I am having a hard time grasping the solution. Thank you
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Assuming the books are distinct, the first one is $15 \cdot 14 \cdot 13 \cdot 12 = 4!\binom{15}4$ as you rightly said.
For (b), you could use multichoose, $\left(\kern-3pt\binom nk\kern-3pt\right)=\binom{n-k+1}{k}$. You are choosing $4$ out of $15$ friends, with repetition allowed. This can be done in $\left(\kern-3pt\binom {15}4\kern-3pt\right)$ ways. But then it depends on how many books you give out. If you're giving them all to one person, there is only one way to do it. If you're giving them to two people, then there are $4!/(2!2!)$ ways to do it, and so on.
So let's do it case by case. If you give them all to one person, you just choose the person, there are $\binom{15}1$ ways to do this.
If you give three to one person, you first choose which three ($\binom43$ ways), then which person ($\binom{15}1$ ways), and then choose the remaining person to give the remaining book ($\binom{14}1$ ways).
The remaining case is two books to two different people. Choose which two books $(\binom42$ ways) to give the first person you pick ($\binom{15}1$ ways), and then choose the remaining person to give the remaining books to ($\binom{14}1$ ways).
Thus the answer is $$\binom{15}1+\binom43\binom{15}1\binom{14}1+\binom42\binom{15}1\binom{14}1=\boxed{2115\text{ ways.}}$$
Luke Collins
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