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A grandma is about to distribute r thousand-dollar banknotes among her n grandchildren.

(a) In how many ways can grandma distribute the money? (Two distributions are the same if every grandchild has the same amount of money.)

(b) Grandma carefully wrote down all the possibilities in which she can distribute the money from (a) and then she picked one such distribution at random and distributed the money accordingly. What is the probability that the oldest grandson Adam gets exactly k banknotes?

(c) What is the probability that every grandchild gets at least some money?

So for a, it's a classic stars and bars problem. So the answer would be $\binom{ r + n - 1 }{ r }$

For b, I think the answer is the number of ways that her oldest grandson gets exactly k banknotes over the total number of ways her grandchildren can receive the banknotes (which derives from a, $\binom{ r + n - 1 }{ r }$ ways). The answer from my teacher is ${\binom{ r-k + n - 2 }{ r-k } \over \binom{ r + n - 1 }{ r }}$. I don't get it. Why is the answer that way? Can somebody help me explain?

For c, I guess it's just a stars and bars with restrictions Stars and Bars with bounds. "every grandchild gets at least some money" = "every grandchild gets at least 1 banknote". So for 1 <=i <=n, denote Xi as each grandchild, the probability needed to find = the number of ways that each grandchild receives at least 1 banknote over the total number of ways grandma distributes the money. Therefore, the answer is ${\binom{ (r-n) + n - 1 }{ r-n } \over \binom{ r + n - 1 }{ r }}$ = ${\binom{ r-1 }{ r-n } \over \binom{ r + n - 1 }{ r }}$ = ${(r-1)!r!\over (r-n)!(r+n-1)!}$ Can somebody help me check?

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For (b), just give the oldest grandson the $k$ banknotes ahead of time.

What is left to distribute? You have $n-1$ remaining grandchildren waiting for their chance at some of the money and $r-k$ remaining money to distribute. Now, apply the result you used for (a) with these new values.

For (c) that is fine, but its more traditional to see the numerator written as $\binom{r-1}{n-1}$ not that it matters. Note that they are equal as $\binom{a}{b}=\binom{a}{a-b}$.

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