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how many numbers greater than 500000 can be made from digits 2,3,4,5,6,and 7 without repeatition if the number is odd?

 Apr 9, 2020
 #1
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2, 3, 4, 5 , 6 , 7

 

 

If  6  is chosen first , we have  3 possibilities for the  last digit (3, 5 or 7)

And we  have 4!  = 24  ways to permute the remaining digits

So  1 * 4! * 3  = 1* 24 * 3  =  72

 

If either  5 or 7 is chosen first, we  have 2 ways to choose  the last digit  and 4!  ways to permute  the remaining digits  =    2 * 4! * 2   =  4 * 24  =  96

[For example....if  we  choose 5 first, we  have two choices for the last digit, either 3 or 7.....and 4!  ways = 24 ways  to permute  the middle four digits  '

 

So....the total possibiities =   72  + 96   =  168

 

 

cool cool cool

 Apr 9, 2020
 #2
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xxxxxxxx

 Apr 9, 2020
edited by Guest  Apr 9, 2020

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