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Let S be the set of permutations of (1,2,3,4,5,6)  whose first term is not 1.  If we choose a permutation at random from S, what is the probability that the third term is equal to 3, and the fourth term is equal to 4?

 Dec 30, 2020
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Possible sets  =

1 can  appear in any of the 5 other positions  and  the other elements can be arranged in 5!  = 120  ways

So   5 *120   =   600  sets

 

In the sets  of interest :

1 can  appear in positions 2, 5 or 6  =   3 ways

And  2, 5  and 6  can be arranged in 3!    = 6 ways in the remaining positions    

So....   3 * 6   =  18  sets

 

Probability  =   18/600  =  3/100

 

 

cool cool cool

 Dec 31, 2020
edited by CPhill  Dec 31, 2020

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