compute \(\binom40+\binom51+\binom62+\binom73.\)
In general: \((\frac{m}{n})=\frac{m!}{n!(m-n)!}\)
so:
\((\frac{4}{0})=1\\ (\frac{5}{1})=5\\ (\frac{6}{2})=15 \\ (\frac{7}{3})=35\)
I’ll leave you to add up the numbers!
(Note that 0! = 1)
Note that
C(n, 0) + C(n+1,1) + C(n + 2, 2) + C(n + 3, 3) +.....+ C(n + m, m) = C(n + m + 1, m)
So
C(4,0) + C(5, 1) + C(6,2) + C(7,3) = C(8,3) = 56
