Question 271307: If you have a 30 digit number and 0-9 are represented equally so there are three of each number how would you calculate how many permutations there are with no repeats so for example: aaabbc if this was a permutation and you were listing all the permutations you wouldn't want to list representations of all three a's at the beginning. so instead of listing aaabbc aaabbc aaabbc you would only list it once. How would you calculate this for the example given at the beginning and how would you calculate it so that 0 wouldn't have been in the first position. So like if you had a 5 digit number 01234 wouldn't happen. I know how to do permutations, so for a 30 digit number it would be 30!, but with these added rules I am having a hard time. Thank you for your time.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! If you have a 30 digit number and 0-9 are represented equally so there are three of each number how would you calculate how many permutations there are with no repeats so for example: aaabbc if this was a permutation and you were listing all the permutations you wouldn't want to list representations of all three a's at the beginning. so instead of listing aaabbc aaabbc aaabbc you would only list it once. How would you calculate this for the example given at the beginning
Ans: 30!/(3!)^10
Example 1122 has only 4!/(2!*2!) = 3*2*! = 6 permutations
They are 1122,2112,2211,1221,1212,2121
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how would you calculate it so that 0 wouldn't have been in the first position.
Ways to put another number in the first position: 9
# of permutations following: 29!/[9!*2!]
Total # of permutations: 9[29!/[9!*2!]
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Cheers,
Stan H.
So like if you had a 5 digit number 01234 wouldn't happen. I know how to do permutations, so for a 30 digit number it would be 30!, but with these added rules I am having a hard time. Thank you for your time.
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