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1 year ago

I have a question here.

Mathematics

1 answer:

aniked [119]1 year ago

5 0

Answer:

3 ways

Step-by-step explanation:

n = 3; r = 2

$nC_{r}=\dfrac{n!}{r!(n-r)!}\\\\3C_{2}=\dfrac{3!}{2!1!}=\dfrac{3*2*1}{2*1}=3$

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2 years ago

I have a question here. ​

I have a question here.