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There are 3 red chips and 2 blue chips. When arranged in a row, they form a certain color pattern, for example RBRRB. How many color patterns are possible?

There are 5 chips in total , hence 5! ways of arranging them. Out of them 3 are similar and 2 are similar, hence divide by 3! and 2! . So the final outcome will be 5! / ( 3! * 2! ) = 10

The pattern depends where the 2 blue chips will be placed in the row (the others will inevitably be red). To place the first blue chip, we have 5 possibilities, for the second one 4, so this would give 5x4=20 possibilities. But there is no difference in pattern regarding the two blue chips (they are identical), therefore, we should divide by 2, so there are 5x4/2=10 different possible patterns. The correct answer is A.
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PhD in Applied Mathematics Love GMAT Quant questions and running.

thanks fluke. that was a mistake on my part.. i meant 2 choices for each slot, ie 5x2=10. but i get that the choices decrease as the r/b balls are used up.. thanks so much!!

Straight forward solution. This is an arrangement of 5 items involving a set of 3 identical items and another of 2 identical items. Solution: 5!/(3!*2!)= 10 ways.

Where the formula is giving the answer to be 10 number of ways to arrange the chip, I am getting 11 ways when arranging them manually. Those 11 ways of arrangements are as follows. BBRRR BRBBR BRBRB BRRBR BRRRB RBBRR RBRBR RBRRB RRBBR RRBRB RRRBB

Can anybody help me on where am I wrong?

Last edited by doe007 on 13 Jun 2012, 05:15, edited 1 time in total.

Where the formula is giving the answer to be 10 number of ways to arrange the chip, I am getting 11 ways when arranging them manually. Those 11 ways of arrangements are as follows. BBRRR BRBBR BRBRB BRRBR BRRRB RBBRR RBRBR RBRRB RRBBR RRBRB RRRBB

There are 3 red chips and 2 blue chips. When arranged in a row, they form a certain color pattern, for example RBRRB. How many color patterns are possible?

This problem can be approached in two ways: 1. Total number of arrangements/(arrangements of R's & S's) = 5!/3!2! = 10 Which is basically removing the arrangements from permutations to get the combinations, or, 2. Total number of ways in which 3 R's can be arranged at 5 places, rest of the places would be filled by B, or 5C3 = 10

Whenever you have a question like this with two identical sets, think of it as 5! where you have a total of 5x4x3x2x1= 120 different sets of rearrangements. But you also have 3 identical red chips and 2 identical blue chips so you MUST account for these as well.

5!/3!2!= 5x4x3x2x1/ 3x2x1x2x1 which equals out to 10 different arrangements being possible.

Would this be the most efficient approach on the GMAT?

Whenever you have a question like this with two identical sets, think of it as 5! where you have a total of 5x4x3x2x1= 120 different sets of rearrangements. But you also have 3 identical red chips and 2 identical blue chips so you MUST account for these as well.

5!/3!2!= 5x4x3x2x1/ 3x2x1x2x1 which equals out to 10 different arrangements being possible.

Would this be the most efficient approach on the GMAT?

THEORY FOR SUCH KIND OF PERMUTATION QUESTIONS:

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!}.

BACK TO THE ORIGINAL QUESTION: There are 3 red chips and 2 blue chips. If they form a certain color pattern when arranged in a row, for example RBRRB, how many color patterns are possible? A. 10 B. 12 C. 24 D. 60 E. 100

According to the above the # of permutations of 5 letters BBRRR of which 2 B's and 3 R's are identical is \frac{5!}{2!*3!}=10.

somebody asked above y would we divide by 3! and 2!.

Its because the 3 reds and 2 blues are identical on their own. if the anagram would have been r(1)r(2)r(3)b(1)b(2) (the numbers in paranthesis is to denote the subscripts for each ) the anagram r(1)r(2)r(3)b(1)b(2) would be diff compared to r(2)r(1)r(3)b(1)b(2) .But here the case is identical .So we have to remove the redundant one's.

There fore arranging them in 5! and later dividing by 3! and 2! to remove the redundant cases.