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Bunuel
How many different four-letter words can be formed (the words need not be meaningful) using the letters of the word GREGARIOUS such that each word starts with G and ends with R?

(A) 8P2

(B) 8P2/(2!*2!)

(C) 8P4

(D) 8P4/(2!*2!)

(E) 10P2/(2!*2!)


BUNUEL ..can you please explain this one?
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GREGARIOUS has 10 letters, but one letter g and one letter r compulsorily taken to be the first letter and last letter respectively.

G _ _ R

so now there are the letters EGARIOUS. These are only 8 letters. None of the letters are repeats. Two of these 8 must be chosen to fill the spaces, but order matters. One of 8 letters can go in the first space. One of 7 letters can go in the second space. So there are 8·7

nPr is n!/(n-r)!, 8P2 = 8!/(8-2)!=8!/6!=8·7

A
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