This is confusing me in Intro to Stats course.. any help in explaining is appreciated!

Binomial Distribution formula:

n!(n−k)!k!\frac{n!}{(n-k)!k!}

For 125 Coin Flips w/ 3 Heads = 317,750 (n = 125, k = 3)

Steps to solve:

125!(125−3)!3!\frac{125!}{(125-3)!3!}

125∙124∙1233∙2∙1\frac{125\bullet124\bullet123}{3\bullet2\bullet1}

1,906,5006\frac{1,906,500}{6} = 317,750

In the steps to arrive at this answer explain that in step 2 125! cancels out (125 – 3)!. I do not understand why

(125∙124∙123125\bullet124\bullet123) cancels (122∙121∙120122\bullet121\bullet120).

Thanks in advance – I hope I articulated my question well enough to get an answer!

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1 Answer

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You’re misinterpreting (n−k)!(n-k)!. What you have is

125!=125⋅124⋅123⋅…⋅3⋅2⋅1125! = 125 \cdot 124 \cdot 123 \cdot \ldots \cdot 3 \cdot 2 \cdot 1

(125−3)!=122!=122⋅121⋅120⋅…⋅3⋅2⋅1(125 – 3)! = 122! = 122 \cdot 121 \cdot 120 \cdot \ldots \cdot 3 \cdot 2 \cdot 1

As you can see, in the first case, you have the product of all positive integers from 1 to 125 (inclusively) and in the second case, all positive integers from 1 to 122. When you take the ratio of the first quantity to the second, you can match all positive integers from 1 to 122, so you are left with 125⋅124⋅123125 \cdot 124 \cdot 123 in the numerator.

Wow! THANK you.. that’s very clear, now!

– Cory Robinson

2 days ago