Question 1195877: How many five-letter permutations can be made from the letters in the word "ADDITION"?
Answer by greenestamps(13198)   (Show Source):
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The 8 letters in the word are 6 different letters, with 2 "D"s and 2 "I"s.
Consider different cases where (1) both the two "D"s and the two "I"s are used, (2) the two "D"s are used, (3) the two "I"s are used, or (4) no letters are repeated.
(1) 2 "D"s and 2 "I"s, and 1 of the other 4 letters:
The letters are DDIIX, where X is one of the other 4 letters.
Choose 1 out of 4 of the other letters in C(4,1) = 4 ways;
The number of different arrangements of DDIIX is (5!)/((2!)(2!)) = 120/4 = 30;
Total number of permutations: 4*30 = 120
(2) 2 "D"s and 3 other all different letters:
The letters are DDXYZ, where X, Y, and Z are 3 different of the other 5 letters.
Choose 3 out of 5 of the other letters in C(5,3) = 10 ways;
The number of different arrangements of DDXYZ is (5!)/(2!) = 120/2 = 60;
Total number of permutations: 10*60 = 600
(3) 2 "I"s and 3 other all different letters:
(same analysis as case (2): 600 permutations)
(4) no repeated letters:
The letters are VWXYZ, which are 5 of the 6 different letters.
Choose 5 of the 6 different letters is C(6,5) = 6 ways;
The number of different arrangements of VWXYZ is 5! = 120;
Total number of permutations: 6*120 = 720
(Note for this case you could also simply do 6 choices for the first letter, 5 choices for the second, ..., 2 choices for the 5th, giving you the answer as 6*5*4*3*2 = 720.)
Total number of permutations for all four cases: 120+600+600+720 = 2040
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