We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
138
3
avatar+42 

 

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

I'm not sure how to solve these binomial expressions.

 Apr 9, 2019
 #1
avatar+5662 
+2

\(\dbinom{n}{k} = \dfrac{n!}{k!(n-k)!}\\ n! = n \cdot (n-1) \cdot (n-2) \cdot \dots \cdot 3 \cdot 2\)

.
 Apr 9, 2019
 #2
avatar+42 
0

Thanks!

detkitten  Apr 9, 2019
 #3
avatar+102417 
+3

Note that

 

C(n,m) + C(n + 1, m + 1) + C(n + 2, m + 2)  + .....+ C( n + q, m + q)  =  C (n + q + 1, m + q)

 

So

 

C(4,0)  + C(5, 1) + C(6,2) + C(7, 3)  = C(8, 3)   =   56

 

 

cool cool cool

 Apr 9, 2019

7 Online Users