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compute \(\binom40+\binom51+\binom62+\binom73.\)

 Mar 10, 2018
 #1
avatar+28132 
+2

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)

 Mar 10, 2018
edited by Alan  Mar 10, 2018
 #2
avatar+103120 
+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

 

 

cool cool cool

 Mar 11, 2018

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