Is there an alternate method to Anagrams?

Question

Hi Folks,
Appreciate help from members here to educate me on using Permutations/Combinations method instead of using ANAGRAMS.

For instance, the number of *distinct* ways you can arrange the word QUANTITATIVE using the anagram method looks like 12!/(2!*3!*2!), (basically 2!,3! and 2! in the denominators being repetitions of letters A,T and I). Though this method is straight forward, questions in GMAT doesn't make this very apparent. Instead we need to apply basic logic, arrive at the ANAGRAM word from question stem and THEN apply the above method. Sometimes I resolve the ANAGRAM word right and sometimes I don't.

So, I was wondering there should be a direct permutation/combination approach to solve word arrangement problems? Any insight will be greatly helpful.

Thanks!

Sharvickr · Answer

Could you please give examples/instances for the same

mniyer · Answer

There can be many examples. Almost all the problems in MGMAT series feature the ANAGRAM approach. But in general, if we are asked to find the number of possible ways to distinctly arrange a word, say "QUANTITATIVE" how do we approach using a perm/comb idea? For the e.g. in question, a part of it sounds permutation (for QUNVE because order does matter) and the rest looks like permutation and combination (because order does not matter within AA, TT or II but it DOES when we do across A,T,I ).

How do we put these pieces together?

Sharvickr · Answer

You could think of it in the following way First a basic example, "AB" can be arranged in 2! {"AB", "BA"} ways. But "AA" can be arranged only in one way For the word "QUANTITATIVE" group in the following manner (QUNVE AA II TTT) AA --> 2! (if letters are distinct) TTT --> 3! (if letters are distinct) II -->2! (if letters are distinct) Since the above three groups have repetitions, out of 10! possible ways, the latter three groups have only one possible way of arrangement. Therefore, no of ways for "QUANTITATIVE" is 10!/(2!*2!*3!) Hope this helps

Bunuel · Answer

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!} .

mniyer · Answer

Great post! Thanks Bunuel!!

Is there an alternate method to Anagrams?

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