Question:

Argue that there are exactly solutions of for which exactly of the are equal to .

My Logic:

I started by interpreting to be the number of ways you can select out of the terms to be zero. But, I can't figure out what \binom{n-1}{n-r+k} does. I think the part is essentially saying you line up ones and create a spot between each one where you can insert a bar. I.e. if you have something like:

1 | 1 | 1 | 1 | 1 | 1 | 1 | 1

Where the |'s are possible places to insert a bar. In which case it seems to me that you would want to choose bars giving you . Can someone please point out to me where my logic is flawed? Thanks.