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JusTLucK04
GMAT 1: 730 Q51 V38
Posts: 270
22 Apr 2014, 07:07
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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How many different anagrams can you make for the word MATHEMATICS?
I donot have any options...but the correct answer here is 8!/2!
I don't understand how?
Should it not be 11!/(2!*2!*2!)
I donot have any options...but the correct answer here is 8!/2!
I don't understand how?
Should it not be 11!/(2!*2!*2!)
22 Apr 2014, 07:22
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JusTLucK04
Your solution is correct. The correct answer is indeed 11!/(2!*2!*2!): the number of arrangement of 11 letters out of which 2 A's, 2 M's and 2 T's are the same.
Similar but much harder question to practice: how-many-words-can-be-formed-by-taking-4-letters-at-a-time-92675.html
THEORY:
Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:
\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).
For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.
Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.
Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).
Hope this helps.
General Discussion
JusTLucK04
22 Apr 2014, 07:28
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Thanks Bunuel..I have been through this post and already using the Math Book and Veritas as my primary source of practice..
This was just to confirm my solution..
Thank you again for the quick replies every time I am stuck some where..
This was just to confirm my solution..
Thank you again for the quick replies every time I am stuck some where..
