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2016-03-31T06:31:44+05:30

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If n is the size of the larger collection, and r is the number of elements that will be selected, then the number of combinations is given by

In the question just posed, n = 20, r = 3, and n – r = 17.  Therefore

To simplify this, consider that:

20 = (20)(19)(18)(17)

That neat little trick allow us to enormously simplify the combinations formula:

= 1140

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@mohitraj. Your answer is wrong
-_- thats half of my things and you copied :P
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2016-03-31T06:32:46+05:30

This Is a Certified Answer

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

If n is the size of the larger collection, and r is the number of elements that will be selected, then the number of combinations is given by


# of combinations =  \frac{n!}{r!(n-r)!}


In the question just posed, n = 20, r = 3, and n – r = 17.  Therefore,


# of combinations =  \frac{20!}{3(17)!}  


To simplify this, consider that:


20! = (20)(19)(18)(17)(the product of all the numbers less than 17)

 

Or, in other words,


20! = (20)(19)(18)(17!)

 

That neat little trick allow us to enormously simplify the combinations formula:


# of combinations =   \frac{(20)(19)(18)(17)}{3!(17)!}  \frac{(20)(19)(18)}{(3) (2) (1)}   \frac{(20)(19)(18)}{(3)(2)(1)}  = 1140

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