With four yellow caps, and seven orange caps, how many ways can eleven boys be provided with caps?
I think this is a permutation problem.
If you order the Yellow and Orange caps in the following order:
Y, Y, Y, Y, O, O, O, O, O, O, O
It is different from:
Y, O, Y, O, Y, O, Y, O, O, O, O
So to me this is a permutation problem.
Apply formula.
11P11 = 39 916 800 ways
I'm not confident in my answer.
+2666 I think it is \(11 \choose 7\). There are 7 ways to choose the boys with orange caps, and the rest have yellow caps.
Thus, there is \(\color{brown}\boxed{330}\) ways. (Note, \({11 \choose 7} = {11 \choose 4}\), so the answer would be the same either way.
