SOLUTION: How many distinct 4 digit numbers can be formed from 1, 1, 2, 2, 2, 3, 4? I have found many problems with one digit repeating that remove the repeated numbers by division, but I c

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Question 838890: How many distinct 4 digit numbers can be formed from 1, 1, 2, 2, 2, 3, 4? I have found many problems with one digit repeating that remove the repeated numbers by division, but I can't figure out what to do with 2 repeating digits aside from lists which are rather inefficient. (7*6*5*4)/(3! +2) is the closest to what I believe to be the answer.
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
1, 1, 2, 2, 2, 3, 4
There are 4 types of 4-digit numbers to get all the permutations of.  These 4
types are ABCD, AABC, AABB, and AAAB.

1.  ABCD There is only one of this form, 1234.  That accounts for 1 form times
4! or 24 ways to arrange it.

2.  AABC We can choose the digit for A 2 ways, For each of those ways, we can
choose the B and C any of C(3,2)=3 ways. That accounts for  2×3 = 6 ways to
choose the digits to rearrange, (They are 1123,1124,1134,2213,2214,2234).  Each
of these 6 can be rearranged in  = 12 ways.  That's 6×12=72 ways.   

3.  AABB.  This is the number of ways to rearrange 1122 which is  = 6 ways.    

4.  AAAB. The digit for A must be 2.  There are then 3 choices of digits for B.
(These are 2221,2223,2224). Each of those can be arranged  = 4 of each
of this form).  That's 12 ways.

Total: 24+72+6+12 = 114 ways.

Edwin

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