+0  
 
0
111
1
avatar

How many 4-digit permutations can be formed out of:1, 1, 1, 2, 2, 3, 4, 5. Thanks for help.

 Jun 6, 2020
 #1
avatar
0

It is much easier to find the "combinations" first and then calculate the permutations from those combinations:

 

There are:  {1, 1, 1, 2}, {1, 1, 1, 3}, {1, 1, 1, 4}, {1, 1, 1, 5}, {1, 1, 2, 2}, {1, 1, 2, 3}, {1, 1, 2, 4}, {1, 1, 2, 5}, {1, 1, 3, 4}, {1, 1, 3, 5}, {1, 1, 4, 5}, {1, 2, 2, 3}, {1, 2, 2, 4}, {1, 2, 2, 5}, {1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5}, {1, 3, 4, 5}, {2, 2, 3, 4}, {2, 2, 3, 5}, {2, 2, 4, 5}, {2, 3, 4, 5} = 22 combinations.

 

Four of them have =4!/3!  x  4       =16 permutations

One of them has    =4!/2!2!            =6 p

Twelve of them have =4!/2!  x 12  =144 p

Five of them have      =4!  x  5       =120 p

 

So: 16  +  6  + 144  + 120 =286 permutations.

 Jun 6, 2020

13 Online Users

avatar