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Question:

A mother has 10 identical coins, and 3 kids.

a) If the first kids gets 3 coins, and the last kid gets 2 coins. How many ways can she spread the rest of the coins?

b) If the second kid only gets 3 coins. How many ways can she spread the coins?

c) If the second kid can have 3 coins or more. How many ways ?

d) If the second kid can have at most 3 coins. How many ways ?

My solution:

a) r = 10 - 3 - 2 = 5 coins leftover

r = 5

n = 3

No order, and repetitions are allowed.

$\therefore$ $n+r-1\choose r$

$\therefore$ $3+5-1\choose 5$ is the answer

b) The second kid is restricted to 3 coins only. Then n = 2, and r = 7.

$\therefore$ $n+r-1\choose r$

$\therefore$ $2+7-1\choose 7$ is the answer

c)

r = 10 - 3 = 7

n = 3 (because the second kid can still receive the rest of the coins)

$\therefore$ $n+r-1\choose r$

$\therefore$ $3+7-1\choose 7$ is the answer

d)

$2+10−1 \choose 10$+$2+9−1 \choose 9$+$2+8−1 \choose 8$+$2+7−1 \choose 7$

Am I right with my logic ?

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    I created this question myself to check my understanding. I want to see if I understand the topic. – 2017-02-23
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    I think your formula should be $\binom{n+r-1}{r-1}$. And I am assuming the coins are identical? – 2017-02-23
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    No the correct formula is $n+r -1\choose r$ or $n+r -1 \choose n -1$. I guess you forgot to place n by r. – 2017-02-23
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    Yes assuming the coins are identical in this question. Do you think my answer is right ? – 2017-02-23
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    Ah, you know what it was; the number of $r$-tuples of non-negative integers whose sum is $n$ equals $\binom{n+r-1}{r-1} = \binom{n+r-1}{n}$. That is usually how $r,n$ are used for this type of problem. But I mistook *your* use of the letters $r,n$. You looking for the number of $n$-tuples of non-negative integers whose sum is $r$, in which case you are correct to exchange the roles of $r,n$ in these binomial coefficients. So I think part $a)$ is correct. But for the part $b)$, are you restricting the second kid to exactly three coins (in which case your work is right) or at most three coins? – 2017-02-23
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    Ok, I see. Yes, I am restricting the second kid to exactly 3 coins, which as you said should be right. However, if the question would ask at most three coins. Then the second kid can still receive the rest of the coins. Thus, it would have been $3+7-1 \choose 7$. – 2017-02-23
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    Thanks for confirming @joeb. Appreciate it. – 2017-02-23
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    If the second kid gets at most three coins, then I believe the answer would be $\binom{2+10-1}{10}+ \binom{2+9-1}{9} + \binom{2+8-1}{8}+\binom{2+7-1}{7}$, allowing for the cases where the second kid gets $0$ coins, $1$ coin, $2$ coins, or $3$ coins. – 2017-02-23
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    Oh, maybe I don't understand the meaning of the word at most. You are right, it means maximum. So it can be 1 coin, 2 coins, or 3 coins given to the second kid. You are right! – 2017-02-23
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    or if they get 0 coins as well – 2017-02-23
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    @joeb , part c question is different from part d. For part c, is the second kid not part of the n = 3 ? I feel like I need to make it look like $2+7-1 \choose 7$ + ... + $2+1-1 \choose 1$ + $2+0-1 \choose 0 $ ? Because the second kid is getting at-least 3 coins. He can 3, 4, and up to 10(max). – 2017-02-23

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