# Thread: x1 + x2 + x3 = 100

1. ## x1 + x2 + x3 = 100

How many solutions in set N ∪ {0} have equation x1 + x2 + x3 = 100.

2. lets fix a value of x1, say 5. How many solutions are left?

The new equation is x2+x3=100-5
Which has 96 solutions. These correspond to x2 taking every value between 0 and 95, and x3=95-x2.

Generalising, we could fix x1 at any value between 0 and 100.
So the number of solutions is:
$\sum_{k=0}^{k=100} 101- k$

Can you finish from there?

3. We can put N identical objects into K distinct cells
in $\binom{N+K-1}{N}$ ways.
Here we have 100 ones.