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# Counting

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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
0

isnt 13/0 error so it doesn't work

Mar 18, 2020
#2
0

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

Guest Mar 18, 2020
#3
0

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

Guest Mar 18, 2020
#4
+485
+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