# $a+b+c=n$ find number of ways [duplicate]

This question already has an answer here:

Please tell me how to find the total number of intergral solutions of $a+b+c=n$ I already know that total number of solutions will be $(n+3−1)c(3−1)$.

but if value of $a$ and $b$ and $c$ is given then what will be the answer??

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## marked as duplicate by Gerry Myerson, Dan Rust, Start wearing purple, azimut, Chris GodsilJul 12 '13 at 12:03

This question was marked as an exact duplicate of an existing question.

if $a,b,c$ are given, then what do you intend to count? – Aang Jul 12 '13 at 6:00
number of ways for example value of a=2 ,b=1 and c=0 so value of n=3 then answer is 3 like (a a b) (a b a) (b a a) sorry for not correct explaination – user85857 Jul 12 '13 at 6:02
It is a combinatorial number. – eccstartup Jul 12 '13 at 6:03
then what will be the formula for above question i am week in permutation and combination – user85857 Jul 12 '13 at 6:06
@user85857: so, your question is: to find the number of ways of selecting $3$ integers from given set of integers so as to make sum $n$ – Aang Jul 12 '13 at 6:06

## 1 Answer

I didn't understand. What means '$a,b,c$ are given' ? I have a solution of other problem:

A question: $n$ is given positive integers. $a, b, c$ are non-negative integers. What is number of $(a,b,c)$ triples of the solutions of the equation $a + b + c = n$

Solution: Let's solve $a+b+c=7$. A solution $a=2,b=4,c=1$. Now, we represent this solution with the symbol $oo/oooo/o$. There are two $/$ and seven $o$. Another solution $a=5,b=2,c=0$. We represent this solution with the symbol $ooooo/oo/$. Again, there are two $/$ and seven $o$. So conclude that there is a one to one corresponding between number of $(a,b,c)$ triples of the solutions of the equation $a + b + c = 7$ with repeated permutations of $ooooooo//$. This number is $\frac{9!}{7!.2!}$ or $C(9,2)$.

Generally, for $a + b + c = n$, number of $(a,b,c)$ triples of the solutions is $C(n-2,2)$

More generally, for $m$ varibles $x_{1} + x_{2} + ... + x_{m} = n$, number of solutions on non-negative integrals is $C(n+m-1,m-1)$

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This doesn't answer the question raised by OP. Of course, he/she didn't clearly state the problem.You can read the comments below the question to get the real problem. – Aang Jul 12 '13 at 9:24
Thanks for your interest. In spite of I read all comments, I didn't understand original problem. If anybody can explain the original problem then, I hope that I will solve the problem. Best regards... – lokman gokce Jul 12 '13 at 9:56