combinations question help Let $N > 10$ be an integer. We have $N$ one-dollar bills that we split between three people. No person can be given less than two one-dollar bills, and everyone gets a whole number of bills (no cents). In how many ways can we do this?
This is how I tried solving the problem. Before splitting, if we give each of the three people two (2) one-dollar bills, then we'll be starting with $N-6$ total bills. Since order doesn't matter, there could be $\displaystyle{{N-6}\choose3}$ ways to do this. Is this right or am I missing something? Any help would be appreciated.
 A: You are right about the first part of your answer. Since each person automatically receives at least $2$ dollars, the problem is equivalent to finding the number of ways $N - 6$ can be split as a sum of three non-negative numbers. I will provide a different method which does not rely on knowing the stars and bars argument. Let $S_3(N - 6)$ be this number. In a similar manner, we may define $S_2(n)$ to be the number of ways in which $n$ may be split as the sum of two non-negative integers, where the order matters. Note that $S_2(n) = n + 1$ for any $n \geq 0.$ It follows that $S_3(N - 6) = \sum_{0 \leq k \leq N - 6} S_2(k) = \sum_{0 \leq k \leq N - 6} (k + 1) = \frac{(N - 5)(N - 4)}{2} = \binom{N - 4}{2}.$ This is because we imagine giving the first person $k$ bills. Then we will be forced to split $N - k$ bills between the remaining two. Generalising this method leads to an interesting discovery. Let us denote by $S_n(p)$ the number of ways in which we can write $p$ as the sum of $n$ non-negative integers. Also define by convention $S_0(0) := 1,$ $S_0(l) := 0$ if $l > 0$ and note that $S_1(l) = 1$ for all $l \geq 0.$ Then by the same reasoning as above, we get that for any $0 \leq k \leq n,$ the following equality holds: $$S_n(p) = \sum_{0 \leq j \leq p} S_{n - k}(p - j) S_k(j) = S_{n - k} \ast S_k (p),$$ where $\ast$ denotes the convolution product. It is somehow interesting to find the convolution product in a problem where it isn't obvious how it might sneak up on us a priori. I hope this helps. :)
A: You're right about starting with figuring out how many ways to give out $N-6$ bills.
The method you need is called the "stars and bars" argument.  Imagine the three people are lined up.  There are two dividers separating people.  There are $N-6$ bills.  Together there are $N-4$ things.  How many ways can you arrange the two dividers among them?  ${N-4 \choose 2}$
A: This problem can be viewed as How many integer solutions which each lager than $2$ of the following integer equation
$$ N_1 + N_2 + N_3 = N $$
where the $N_i$ is the number of one-dollar bill that the i-th people get, so what we want is find integer solutions with $N_i \ge 2$ , we can simplify the problem further in the following:
$$ (N_1-2) + (N_2-2) + (N_3-2) = N-6 $$
rewrite as following with denoting $x_i = N_i - 2$
$$ x_1 + x_2 + x_3 = N-6 $$ where $x_i$ is nonegative integer.
what we want to solve is a classic problem , The Number Of Non-negative Integer Solutions Of Equations
Using the proposition $2$ of the above link ,There are $n+r-1\choose r-1$ distinct non-negative integer-valued vectors $(x_1, x_2,...,x_r)$ satisfying the equation $$ x_1 + x_2 + ... + x_r = n$$
Apply it to this question , we get the answer $N-6 + 3 -1 \choose 3-1 $ and that
is $N-4 \choose 2 $
