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The NYPD has 35,000 members. Suppose two members are chosen at random. In how many different ways can they be chosen?

 

1,225,000,000 pairs

70,000 pairs

1,190,000,000 pairs

1,224,965,000 pairs

 Oct 18, 2017
 #1
avatar+129852 
+1

 

We want to choose any 2 of the 35,000

 

So  

 

C(35000, 2)  =  612,482,500  ways (pairs)

 

Sorry....footballgirl....it doesn't appear that any of the choices you have been given are correct....

 

 

cool cool cool

 Oct 18, 2017
 #2
avatar+2489 
+4

Sir CPill’s answer is right. Normally, in selecting combinations the order of selection is irrelevant.  (XY is the same as YX).

 

However

 

P(35000, 2) = 1,224,965,000

 

This implies the enumeration counts the order of selections in addition to the combinations. In other words, order matters.   The question should state that, if that is the case. If it does not, then none of the answer choices is correct. You should let your teacher know. 

GingerAle  Oct 18, 2017
 #3
avatar+129852 
0

Thanks, GA....didn't think of a permutation.....!!!

 

cool cool cool

 Oct 18, 2017

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