Different ways of giving away 35 coins to 5 people? The first part of the problem asks how many ways there are to give away 35 identical coins to 5 people, and I've concluded that it's ${35 \choose 5}$ because you're selecting how many ways you can give 35 coins to different people. However, I have no idea how to approach this problem when the coins are distinct. If the coins are distinct that means this turns into a permutations problem (because order now matters), correct? So would it be P(35,5)?
 A: When the coins are identical, all that matters is how many coins each person gets. So you are interested in the number of solutions to the equation $a+b+c+d+e=35$, where $a,b,c,d,e \geq 0$ are integers. The classical way of solving this equation is to imagine your 35 coins in a line, and four "separators" that separate the lot of each person. In total there are 39 objects, 35 of one type and 4 of another, so the total number of solutions is $\binom{39}{4}$.
If the coins are distinct, then you can describe each way by saying which person gets which coin. Each coin could go to one of five different people, and there are 35 coins, so in total there are $5^{35}$ different ways.
A: You ought to be more careful. ${35 \choose 5}$ is choosing $5$ items out of $35$ distinct items. You want to split up $35$ indistinguishable items into $5$ groups.
In general, the number of ways to give $n$ distinguishable items to $k$ people is ${n + k - 1 \choose n}$.
Proof. This number is the same as the number of bit-strings with $n$ amount of $0$s and $k-1$ amount of $1$s. Indeed, suppose we had $8$ things and $4$ people. Then take such a bit-string:
$$00100010100$$
The first person gets the number of items equal to the number of zeros before the first $1$, in this case $2$ items. The second gets three items, the third gets one, and the last gets the remaining two.
