+0  
 
0
39
4
avatar

Find the value of the sum \(\dbinom{13}{0}+\dbinom{13}{1}+\dbinom{13}{2}+\dbinom{13}{3}+\dbinom{13}{4}+\dbinom{13}{5}+\dbinom{13}{6}.\)

 Mar 18, 2020
 #1
avatar
0

isnt 13/0 error so it doesn't work

 Mar 18, 2020
 #2
avatar
0

I don't think so, because 0! is one

Guest Mar 18, 2020
 #3
avatar
0

forgive me if i am wrong but i dont see the !

Guest Mar 18, 2020
 #4
avatar+402 
+1

Ok let's use some counting knowledge:

 

(N c 0) + (N c 1) + (N c 2) . . . .. . . . . . .  (N c N) is equal to 2^n    (If you don't believe me, you can try it out yourself). By that logic, (13c0) + (13c1) ......(13c13) = 2^13. Now, how does this relate? That's because realize that (13c0) + (13c1)......(13c6) is equal to (13c7) + (13c8)......(13c13) by symmetry (Think about the definition of N choose k). Therefore, the solution is simply 1/2 * 2^13, which gives us 2^12, and thus an answer of 4096

 Mar 18, 2020

19 Online Users

avatar
avatar
avatar