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$X_1 + X_2 + X_3 + X_4 + X_5 = 10$.

(i) How many non-negative integer solutions are there?

this is the easy part

(ii) How many non-negative integer solutions are there such that $X_1 = X_2$?

Do i just divide the answer in $1$ by $2$?

(iii) How many non-negative integer solutions are there such that $X_1 > X_2$?

I am not sure how to begin

Thanks in advance

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  • $\begingroup$ To answer, I need to know how you are doing the first part. $\endgroup$ – G-man Dec 20 '15 at 14:24
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    $\begingroup$ Look up stars and bars $\endgroup$ – Shailesh Dec 20 '15 at 14:37
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    $\begingroup$ for question 1, my answer is 14C9 $\endgroup$ – soulless Dec 20 '15 at 14:39
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    $\begingroup$ @soulless When you pose a question here, you should include the work you have done in the statement of the problem. $\endgroup$ – N. F. Taussig Dec 20 '15 at 14:48
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    $\begingroup$ It seems to be $14\choose10$ instead. $\endgroup$ – Element118 Dec 20 '15 at 14:50
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i) By stars and bars, there are $\binom{14}{4}=\color{red}{1001}$ ways of writing $10$ as a sum of $5$ non-negative integers.

If $X_1=X_2$, such value can range from $0$ to $5$, and by the same principle as above the answer to ii) is given by $\binom{12}{2}+\binom{10}{2}+\ldots+\binom{2}{2}=\color{red}{161}$. (We have just counted the ways of writing $10-2X_1$ as $X_3+X_4+X_5$).

Half the $1001-161=840$ cases in which $X_1\neq X_2$ are such that $X_1>X_2$, by symmetry. It follows that the answer to iii) is given by $\color{red}{420}$.

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  • $\begingroup$ (+1) You might like to add a line that for $(ii)$ you are counting what remains for $X_3+X_4+X_5$ $\endgroup$ – true blue anil Dec 20 '15 at 17:11

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