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How many different positive three-digit integers can be formed using only the digits in the set {2, 3, 5, 5, 5, 6, 6, 7, 8} if no digit may be used more times than it appears in the given set of available digits?

 Jun 25, 2021
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since there are 9 numbers and choose 3 to be in so we can do 9!/6!=504 but some numbers are repeated in the list which are 5 and 6. 5 is repeated three times and 6 is reapeated twice so we divide by 2! and 3!

504/2/6=42

coolcoolcool

 Jun 25, 2021
edited by Wsai12  Jun 25, 2021

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