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How do you evaluate 8C3?

1 Answer

Aug 3, 2016

#""_8C_3=56#.

Explanation:

We know that, #""_nC_r=(n!)/{(n-r)!*r!}#

With, #n=8, r=3, ((n-r)!)=5!#

So, #""_8C_3=(8!)/(5!*3!)=(cancel(5!)*6*7*8)/(cancel5!*3!)#

#=(6*7*8)/(1*2*3)=56#.

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