how many different string can be made from the word PEPPERCORN when all letters are used and such strings do not contain the substring CON?
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im interested in seeing how this one is done!
no of different strings that can be made from the word PEPPERCORN are \[10!/(3!*2!)\] as there are 10 letters with one letter being repeated thrice and one leter being repeated twice
No of strings with sub string CON are \[8!/(3!*2!)\]
as there are 8 letters (consider whole CON as one letter or unit) and P repeated thrice and E repeated twice
Hence no of different strings without the substring CON are \[10!/(3!*2!) - 8!/(3!*2!)\]
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okay i get it :) thanks... 3! and 2! is from the letters which are repeated right
yup...10! is assuming all the letters are different but our word has three P's which when interchanged do not change the arrangement but have been in included in the 10! as different arrangements...now no of times the each unique arrangement is reapeated is equal to no of times the 3 P's have been interchanged or permuted among themselves. That is 3! hence divide by 3!. Like wise 2! for E's
and why must be 8!, i know 8 is from the prob of CON in string with length 10, and we substract it..hm..why?
@sumanth4phy There are two letter Rs in PEPPERCORN.
Imagine gluing all the letters of CON togetehr as single unit. Now we have the letters "P, E, P, P, E, R, CON" we should not count CON as three letters but one as we need permutations where CON is clubbed hence 8!
CON has to exist as substring which implies u cant treat C, O, N as individual letters any more. There are fixed with respect to each other only the clubbed substring can be shifted here and there with other letters
yeah, i found 2 letters which are repeated twice, R and E, then 1 letter which is repeated thrice, it's P
so 8!/(3!2!2!) or 8!/(3!2!) ?
yup I have overlooked R's solution is \[10!(3!*2!*2!) - 8!/(3!*2!*2!)\]