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anonymous
 3 years ago
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?
anonymous
 3 years ago
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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im interested in seeing how this one is done!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no 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 \[10!/(3!*2!)  8!/(3!*2!)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where did you get the (3!*2!)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay i get it :) thanks... 3! and 2! is from the letters which are repeated right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yup...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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and why must be 8!, i know 8 is from the prob of CON in string with length 10, and we substract it..hm..why?

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.0@sumanth4phy There are two letter Rs in PEPPERCORN.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Imagine 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, 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!) ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yup I have overlooked R's solution is \[10!(3!*2!*2!)  8!/(3!*2!*2!)\]
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