probability problem solution with full proof

Question

n+4 apples are distributed at random among n persons. Find the probability that atleast one of them will receive none.

I want a detailed solve using complex combinatorial summation. Please help me to guide through with proofs

Answer 1

The number of ways to distribute $n+4$ apples among $n$ people so that at least one person gets no apples is the number of ways to distribute the apples, minus the number of ways to distribute the apples so that each person gets at least one apple. I take the question to mean that we only care about how many apples each person gets, not which apples he gets, so we have the problem of distributing $n+4$ indistinguishable objects among $n$ persons which is $$\binom{n+4 + n-1}{n-1} = \binom{2n+3}{n-1}$$. This is the "balls and walls" or "stars and bars" formula.

To figure out how many ways to distribute the apples so that everyone gets at least one apple, just give everyone an apple, then distribute the remaining 4. This is again a "balls and walls" problem, and the answer is $$\binom{4+n-1}{n-1} = \binom{n+3}{n-1},$$ so the final answer is $$\binom{2n+3}{n-1}-\binom{n+3}{n-1}.$$