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How many $3$ letter words can be formed from the word "$TESTBOOK$" ?


Letters in this word are : $T,E,S,B,O,K$

  • Make a $3$ letter word from $6$ distinct => $P(6,3) = 120$ ways
  • Now, Make a $3$ letter word from repeating letters => $C(2,1) * C(3,2) * C(5,1) = 30$ ways

Total $3$ letter words = $150$


Am I right here or missing any possibility ?

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  • $\begingroup$ For the second bullet point: Why not: there are 6 choices for the first letter then 6 choices for the second letter and then 6 choices for the third letter giving $6^3$ choices? It is suspicious that the number of words selected by repetition allowed is les than the correspondig number (with no repetition) $\endgroup$ – zoli Jan 26 '17 at 15:04
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    $\begingroup$ Your answer is correct, but the second formula should be $C(2,1)*C(3,2)*C(5,1) = 30$ $\endgroup$ – true blue anil Jan 26 '17 at 15:06
  • $\begingroup$ @trueblueanil typo corrected :-) $\endgroup$ – Jon Garrick Jan 26 '17 at 15:10
  • $\begingroup$ @zoli there are two sets of characters which may repeat and they may only repeat once. Hence using $O,O$ or $T,T$ as two of the letters in one of our three letter words leaves few possibilities for the third letter. Not to mention we have many more distinct letters than we have letters that repeat. $\endgroup$ – ClownInTheMoon Jan 26 '17 at 16:03
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    $\begingroup$ No, "words ... formed" means permissible permutations of letters, and $C(2,1)*C(3,2)*C(5,1)$ already takes care. (Choose letter that repeats)$\times$ (Choose $2$ places for them)$\times$(Choose another letter and place it in vacant slot), e.g. $TET, TTE, ETT$ are all taken care of. $\endgroup$ – true blue anil Jan 27 '17 at 3:48
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Hint -

For second case take repeating letters as group. So we have 2 groups and 4 distinct alphabets.

Pick 1 from group and 1 from 4 distinct letters.

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