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What is the number of integers greater than 6000 that can be formed using the digits 3,5,6,7 and 8 without repetition?

 Dec 27, 2019
 #1
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You can make: 5! =120 integers > 6000 such as: {3, 5, 6, 7, 8} | {3, 5, 6, 8, 7} | {3, 5, 7, 6, 8} | {3, 5, 7, 8, 6} | {3, 5, 8, 6, 7} | {3, 5, 8, 7, 6} | {3, 6, 5, 7, 8} | {3, 6, 5, 8, 7} | {3, 6, 7, 5, 8} | {3, 6, 7, 8, 5} | {3, 6, 8, 5, 7} | {3, 6, 8, 7, 5} | {3, 7, 5, 6, 8} | {3, 7, 5, 8, 6} | {3, 7, 6, 5, 8} | {3, 7, 6, 8, 5} | {3, 7, 8, 5, 6} | {3, 7, 8, 6, 5} | {3, 8, 5, 6, 7} | {3, 8, 5, 7, 6} | ... (total: 120)

 

P.S. You have 5 numbers, so ANY combination will be > 6000 !!. Unless you meant > 60,000 then: 120 / 5 = 24 integers that start with each digit. So, if you remove 3 and 5 from the beginning of each integer, then you will have:  120 - [2 x 24] = 72 integers > 60,000.

 Dec 27, 2019
 #2
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Three choices for first integer    6   7    8

   second integer has 4 choices of those remaingn on the list

     third integer has 3 choices

      last integer 2 cohoices

 

3 x 4 x 3 x 2  = 72 integers >6000

 Dec 27, 2019

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