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A team has to design a flag. The team has six blue strips of
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22 Dec 2010, 08:21
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A team has to design a flag. The team has six blue strips of cloth and seven red strips of cloth that they must use as is to make the flag. How many different flags can the team design with the materials at hand? (A) 42 (B) 120 (C) 720 (D) 1,716 (E) 5,040
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22 Dec 2010, 08:33



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22 Dec 2010, 08:37
Bunuel which formula is this ? why do you divide 13! by 7! and 6! can you explain that thanks
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22 Dec 2010, 08:41
rxs0005 wrote: Bunuel
which formula is this ?
why do you divide 13! by 7! and 6!
can you explain that
thanks Sure. 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 \(\frac{9!}{4!3!2!}\). So, number of permutation of 13 strips out of which 7 are blue and 6 are red would be \(\frac{13!}{7!6!}\). Hope it's clear.
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Re: Counting PS
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22 Dec 2010, 08:48
Thanks , i was not aware of this formula also here does the order matter ? ( i am assuming yes because its permutation and not combination )
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22 Dec 2010, 10:38
Thanks for the debreifing I read these books and they have this n things can be arranged n! ways eg : ABA can be arranged 3! = 6 ways  1 so ABA can be also arranged in 3!/2! = 3 ways  2 what is the diff between 1 and 2 i am guessing 1 is wrong it cant be 6 ways ? just trying to get the basic clear here thanks
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22 Dec 2010, 10:58
rxs0005 wrote: Thanks for the debreifing I read these books and they have this n things can be arranged n! ways
eg :
ABA can be arranged 3! = 6 ways  1
so ABA can be also arranged in 3!/2! = 3 ways  2
what is the diff between 1 and 2
i am guessing 1 is wrong it cant be 6 ways ?
just trying to get the basic clear here
thanks Permutations (arrangements) of \(n\) distinct things is \(n!\). So, 3 distinct letters A, B, C can be arranged in 3!=6 ways: ABC, ACB, BAC, BCA, CAB, and CBA; 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: \(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\). So, 3 letters AAB, of which 2 are alike, can be arranged in 3!/2!=3 ways: AAB, ABA, and BAA. That's because arrangements of 2 A's don't give different group or in another way AB can be arranged as AB or BA but AA can only be arranged in one way AA. Hope it's clear.
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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.
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Re: Counting PS
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23 Dec 2010, 13:19
Bunuel wrote: Sure.
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 \(\frac{9!}{4!3!2!}\).
So, number of permutation of 13 strips out of which 7 are blue and 6 are red would be \(\frac{13!}{7!6!}\).
Hope it's clear.
this is a great explanation on permutations, would it be possible to add it to the GMAT math book?



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A team has to design a flag. The team has six blue strips of
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08 Feb 2016, 19:23
oh damn, I hate this kind of problems.. first...we have 6 identical blues and 7 identical reds. now, we have 13 totals. thus, we can arrange in 13! ways. since we have 6 blue and 7 red, we need to divide 13! by 6! and by 7!. we get: 13*12*11*10*9*8*7!/7!*1*2*3*4*5*6 simplify: 13*12*11 = 1,726.
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