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Mathematics
(7C0 + 7C1) + (7C2 + 7C3) + .... + (7C6 + 7C7) =
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Q. $\left(^{7}C_{0} + ^{7}C_{1}\right) + \left(^{7}C_{2} + ^{7}C_{3}\right) + .... + \left(^{7}C_{6} + ^{7}C_{7}\right) = $
KEAM
KEAM 2018
A
$2^8 - 2 $
50%
B
$2^7 - 1 $
0%
C
$2^7$
33%
D
$2^8 - 1 $
17%
E
$2^7 - 2 $
17%
Solution:
$\left({ }^{7} C_{0}+{ }^{7} C_{1}\right)+\left({ }^{7} C_{2}+{ }^{7} C_{3}\right)+...+\left({ }^{7} C_{6}+{ }^{7} C_{7}\right)$
$={ }^{7} C_{0}+{ }^{7} C_{1}+\ldots+{ }^{7} C_{7}$
$= 2^{7}\left[\because C_{0}+C_{1}+C_{2}+... +C_{n}=2^{n}\right]$