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I've got the question of how many words can be made from DISKRET without the new word containing SIK, DIS or RET. For example DSIKRTE is not allowed.

My calclation is

7! - 7!/3!-5!-5!-5! = 4560

I have been given the answer 4692 and can't figure out where it goes wrong. Please help.

1 Answers1

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There are $7!$ total permutations. Of those, $5!$ contain the word SIK, $5!$ contain the word DIS and $5!$ contain the word RET.

Notice that no word can contain both SIK and DIS but there are words that contain SIK or DIS and RET. There are $2*3!$ of these words.

By inclusion-exclusion the answer is: $$ 7! - 3*5! + 2*3!=4692$$

Zanzag
  • 1,519
  • Thank you for a very clear answer. I was going down the rabbit hole with different permutations. –  Dec 04 '23 at 18:14