+0

0
40
2

compute $$\binom40+\binom51+\binom62+\binom73.$$

Guest Mar 10, 2018
Sort:

#1
+26547
+1

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)

Alan  Mar 10, 2018
edited by Alan  Mar 10, 2018
#2
+84168
+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

CPhill  Mar 11, 2018

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