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

• Dec 11th 2011, 11:22 AM
nukenuts
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.

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 $\binom{10+6-1}{10}$.

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

$\sum_{n=0}^9 {n+6-1\choose n}$
• Dec 11th 2011, 11:34 AM
TheChaz
Re: No. of solutions for x1 + x2 + x3 + x4 + x5 + x6 < 10
"10 identical items..." - ??? What makes you think that they are identical?

Take a look at this:
number theory - Non-negative integral solutions of $X_1+X_2+X_3+X_4<n$ - Mathematics - Stack Exchange
• Dec 11th 2011, 11:48 AM
nukenuts
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.

• Dec 11th 2011, 11:48 AM
Plato
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.
I would view this as having 10 identical items to distribute among 6 people and would use $\binom{10+6-1}{10}$.
$\color{blue}\sum_{n=0}^9 {n+6-1\choose n}$