Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 06 Apr 2011
Posts: 42

Is there an alternate method to Anagrams? [#permalink]
Show Tags
19 Oct 2013, 20:09
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!



Intern
Joined: 25 Jul 2013
Posts: 4
GMAT 1: 650 Q45 V35 GMAT 2: 700 Q48 V37

Re: Is there an alternate method to Anagrams? [#permalink]
Show Tags
19 Oct 2013, 20:29
mniyer wrote: 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 Could you please give examples/instances for the same



Intern
Joined: 06 Apr 2011
Posts: 42

Re: Is there an alternate method to Anagrams? [#permalink]
Show Tags
20 Oct 2013, 03:06
Sharvickr wrote: mniyer wrote: 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 Could you please give examples/instances for the same 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?



Intern
Joined: 25 Jul 2013
Posts: 4
GMAT 1: 650 Q45 V35 GMAT 2: 700 Q48 V37

Re: Is there an alternate method to Anagrams? [#permalink]
Show Tags
20 Oct 2013, 03:33
mniyer wrote: Sharvickr wrote: mniyer wrote: 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 Could you please give examples/instances for the same 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? 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



Math Expert
Joined: 02 Sep 2009
Posts: 39581

Re: Is there an alternate method to Anagrams? [#permalink]
Show Tags
20 Oct 2013, 04:24
mniyer wrote: 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! 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 \(\frac{9!}{4!3!2!}\).
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 06 Apr 2011
Posts: 42

Re: Is there an alternate method to Anagrams? [#permalink]
Show Tags
20 Oct 2013, 08:47
Bunuel wrote: mniyer wrote: 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! 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 \(\frac{9!}{4!3!2!}\). Great post! Thanks Bunuel!!




Re: Is there an alternate method to Anagrams?
[#permalink]
20 Oct 2013, 08:47







