Question 184193This question is from textbook saxon algebra 2
: ALASKA has 6 letters with A repeated three times. Find the number of distinguishable permutations of letters in ALASKA.
I worked it out to: 6/3=2. Am I correct?
This question is from textbook saxon algebra 2
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! ALASKA has 6 letters with A repeated three times. Find the number of distinguishable permutations of letters in ALASKA.
I worked it out to: 6/3=2. Am I correct?
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How can it be 2? You can list ALASKA, ALKAAS, ALSAAL, SAKALA, SAALAK, etc. That's 5, so it can't be 2.
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Is it using all 6 letters? You didn't say, but if it is:
For the 1st letter, you can choose 1 of 6, then 1 of 5, then 1 of 4, etc, so you get 6*5*4*3*2 = 720 (6 factorial = 6!)
But any choice of a letter A is the same as the other 2, so you have to divide by 3*2*1 = 6.
720/6 = 120
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