No. of solutions for x1 + x2 + x3 + x4 + x5 + x6 < 10

Find the number of non-negative integer solutions of the inequality

x1 + x2 + x3 + x4 + x5 + x6 < 10.

Answer:

If this was -

Find the number of non-negative integer solutions of the equation

x1 + x2 + x3 + x4 + x5 + x6 **=** 10

I would view this as having 10 identical items to distribute among 6 people and would use $\displaystyle \binom{10+6-1}{10}$.

However Im not sure how to handle it with when it's an equality. Would I write it as a summation formula?

$\displaystyle \sum_{n=0}^9 {n+6-1\choose n}$

Re: No. of solutions for x1 + x2 + x3 + x4 + x5 + x6 < 10

Re: No. of solutions for x1 + x2 + x3 + x4 + x5 + x6 < 10

They are identical in that they I am viewing this as having ten 'ones' that are being distributed among 6 people....

Taking x1 + x2 + x3 + x4 + x5 + x6 = 10

So x1 could have 5 'ones' and x2 - x6 could have one 'one' each. Which would be a solution for the equation.

Am I going about this the wrong way with that summation formula in my first post?

Re: No. of solutions for x1 + x2 + x3 + x4 + x5 + x6 < 10

Quote:

Originally Posted by

**nukenuts** Find the number of non-negative integer solutions of the inequality

x1 + x2 + x3 + x4 + x5 + x6 < 10.

Answer:

If this was -

Find the number of non-negative integer solutions of the equation

x1 + x2 + x3 + x4 + x5 + x6 **=** 10

I would view this as having 10 identical items to distribute among 6 people and would use $\displaystyle \binom{10+6-1}{10}$.

However Im not sure how to handle it with when it's an equality. Would I write it as a summation formula?

$\displaystyle \color{blue}\sum_{n=0}^9 {n+6-1\choose n}$

** Your solution is absolutely correct. Way to go.**