How do you evaluate 8C2?

Question

autormtak0w · Accepted Answer

Explanation:  We have, nCr=n!(n−r)!r!  ∴8C2=8!(8−2)!⋅2!=8!6!⋅2!=6!⋅7⋅86!⋅2!=7⋅82!=561⋅2=28

Barbara Meeker · Accepted Answer

In the given problem, we have to evaluate

8 C_{2}

Here, we will use the formula of combination which is

n C_{r} = \frac{n!}{(n - r)! r!}

You can easily identify that n is 8 and r is 2 here.

\Rightarrow 8 C_{2} = \frac{8!}{(8 - 2)! 2!}

= \frac{8!}{6! 2!}

= \frac{8 \times 7 \times 6!}{6! 2!}

After calculating the above e\times pression, we get

= \frac{8 \times 7}{2!}

= \frac{8 \times 7}{2 \times 1}

= \frac{56}{2}

= 28

Therefore,

8 C_{2} = 28

Vasquez · Accepted Answer

The formula for a combination of choosing r unique ways from n possibilities is:nCr=n!r!(n−r)!where n is the number of items and r is the unique arrangements.Plugging in our numbers of n=8 and r=2, we get:8C2=8!2!(8−2)!Remember from our factorial lesson that n!=n⋅(n−1)⋅(n−2)⋅…⋅2⋅1Calculate the numerato r n!:n!=8!K8!=8×7×6×5×4×3×2×18!=40,320Calculate the first denominator (n−r)!:(n−r)!=(8−2)!(8−2)!=6!=6×5×4×3×2×16!=720Calculate the second denominator r!:r!=22!=2×12!=2Calculate our combination value nCr for n=8 and r=2:8C2=40,3202×7208C2=40,32014408C2=28

How do you evaluate 8C_2?

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