+0  
 
0
144
2
avatar

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

Guest Mar 10, 2018
 #1
avatar+26973 
+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
avatar+88898 
+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

CPhill  Mar 11, 2018

31 Online Users

avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.