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How do you evaluate 8C3?

1 Answer

Aug 3, 2016

$_{8} C_{3} = 56$ $""_8C_3=56$ .

Explanation:

We know that, $_{n} C_{r} = \frac{n!}{(n - r)! \cdot r!}$ $""_nC_r=(n!)/{(n-r)!*r!}$

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

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

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

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