# Number of non-negative integers solutions of $x_1 + x_2 + x_3 + x_4 + x_5 = 10$ when $x_1 = x_2$ and when $x_1 > x_2$

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

(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

• To answer, I need to know how you are doing the first part. – G-man Dec 20 '15 at 14:24
• Look up stars and bars – Shailesh Dec 20 '15 at 14:37
• for question 1, my answer is 14C9 – soulless Dec 20 '15 at 14:39
• @soulless When you pose a question here, you should include the work you have done in the statement of the problem. – N. F. Taussig Dec 20 '15 at 14:48
• It seems to be $14\choose10$ instead. – Element118 Dec 20 '15 at 14:50

## 1 Answer

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 $\color{red}{420}$.

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