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Using the digits 2, 3, 4, 7 and 8, Carlos will form five-digit positive integers. Only the digits 2 and 3 can be used more than once in any of Carlos’ five-digit integers. How many distinct fivedigit positive integers are possible?

 Mar 7, 2022
 #1
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In cases such as this one, it is often a lot easier to compute the combinations first, then convert them into permutations as follows:

 

(22222, 22223, 22224, 22227, 22228, 22233, 22234, 22237, 22238, 22247, 22248, 22278, 22333, 22334, 22337, 22338, 22347, 22348, 22378, 22478, 23333, 23334, 23337, 23338, 23347, 23348, 23378, 23478, 33333, 33334, 33337, 33338, 33347, 33348, 33378, 33478) >>Total = 36 combinations.

These 36 combinations break down as follows:

 

2 Ccombinations  x   1 Permutation each == 2P

8 C x 5P ==40P    

2 C x 10P ==20P  

12 C x 20P ==240P     

3 C x 30P ==90P  

8 C x 60P ==480P  

1 C x 120P==120P  

Total combinations ==  36

Total permutations == 992

 Mar 7, 2022
 #2
avatar+36433 
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Start with not using   478

   then you have  2 x 2 x 2 x 2 x 2 = 32   combos

 

then you can have  ONE  8      or  ONE 7     or  ONE  4      each in one of the 5 positions = 15     the other 4 positions 2 x 2 x 2 x 2   possible

                                 15  x 16 = 240

 

then you can have TWO of the 4 7 8       47    74    78   87   48   84             5 C 2   x 6 = 60

     the other three positions  would be   2 x 2 x 2 = 8 

           60 x 8 = 480 

 

you can have all of 4 7 8    in 6 orders   x   5 C3 = 60    the other two positions filled by 2 x 2

                                       4 x 60 = 240

 

 

Summing the reds =      992  possible numbers         ( same answer guest found !)

 Mar 7, 2022

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