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Re: In how many different ways can the letters A, A, B [#permalink]

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26 Jun 2013, 15:10

1

This post received KUDOS

Possible arrangements: 1) xcxxxxxx ---- 1C1 * 6!/(2!*3!) (D can be placed in 1 position) 2) xxcxxxxx ---- 2C1 * 6!/(2!*3!) (D can be placed in 2 positions) 3) xxxcxxxx ---- 3C1 * 6!/(2!*3!) (D can be placed in 3 positions) 4) xxxxcxxx ---- 4C1 * 6!/(2!*3!) (D can be placed in 4 positions) 5) xxxxxcxx ---- 5C1 * 6!/(2!*3!) (D can be placed in 5 positions) 6) xxxxxxcx ---- 6C1 * 6!/(2!*3!) (D can be placed in 6 positions) 7) xxxxxxxc ----- 7C1 * 6!/(2!*3!) (D can be placed in 7 positions)

Adding them all = (1+2+3+4+5+6+7) * 6!/(2!*3!) = 28 * 6!/(2!*3!) = 1680

Re: In how many different ways can the letters A, A, B [#permalink]

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26 Jul 2014, 21:08

Hello from the GMAT Club BumpBot!

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Re: In how many different ways can the letters A, A, B [#permalink]

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23 Sep 2014, 11:00

BarneyStinson wrote:

walker wrote:

A

1) the total number of arrangements: 8! 2) excluding double counting (A1, A2 and A2, A1 are the same): 8!/2!*3! = 3360 3) the number of arrangements with C D is equal the number of arrangements with D C. Therefore, answer is 3360/2 = 1680.

Can you be more clear in your explanation with the step 3?

I considered C to the right of D, the combination together as one unit and there are 7 units to be arranged with 2 A's and 3 B's. Obviously, I was not even close to any of the options. What's wrong with my approach?

I don't think that by being on the right of D they mean C has to be right next to it. C can be anywhere. Therefore you can't "tie" them together and make it into 7 slots. You have to consider all 8 slots. Hope I helped!

Re: In how many different ways can the letters A, A, B [#permalink]

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15 Oct 2015, 10:51

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In how many different ways can the letters A, A, B [#permalink]

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10 Feb 2016, 21:46

Bunuel wrote:

tania wrote:

In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the letter C must be to the right of the letter D? A.1680 B.2160 C.2520 D.3240 E.3360

Can someone explain how I should approach to solve the above problem?

We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\).

Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \(\frac{3360}{2}=1680\)

Answer: A.

Though understood the question, i could not under stand the half left and half right funda. would you please elaborate.?

In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the letter C must be to the right of the letter D? A.1680 B.2160 C.2520 D.3240 E.3360

Can someone explain how I should approach to solve the above problem?

We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\).

Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \(\frac{3360}{2}=1680\)

Answer: A.

Though understood the question, i could not under stand the half left and half right funda. would you please elaborate.?

Re: In how many different ways can the letters A, A, B [#permalink]

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11 Feb 2016, 00:29

robu wrote:

Bunuel wrote:

tania wrote:

In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the letter C must be to the right of the letter D? A.1680 B.2160 C.2520 D.3240 E.3360

Can someone explain how I should approach to solve the above problem?

We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\).

Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \(\frac{3360}{2}=1680\)

Answer: A.

Though understood the question, i could not under stand the half left and half right funda. would you please elaborate.?

Re: In how many different ways can the letters A, A, B [#permalink]

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01 Jul 2016, 22:57

Bunuel wrote:

anilnandyala wrote:

We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: .

Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be

CAN ANYONE EXPLAIN LAST STEP

THANKS IN ADVANCE

Obviously C and D can have ONLY TWO positions: C to the right of C OR to the left, how else?

Now, why should C (or D) be in more cases to the right (or to the left) of D (C)? Does probability favors either of these letters? No. Hence exactly in half of these cases D will be to the right of C and in half of these cases to the left.

Hope it's clear.

I didn't understand the statement "Hence exactly in half of these cases D will be to the right of C and in half of these cases to the left." Asked myself Why? How? (similar situation with the arrangement of Frankie and Joe - question) The following helped me understand: The letters A, B and C can be arranged in 3 slots in 3*2*1 ways. ABC, ACB, BAC, BCA, CAB and CBA. Under normal circumstances C comes before B thrice; B comes before C thrice. For a specific case we consider whatever is required.

I hope my understanding is right. If yes, I hope it helps someone.

-Arvind

gmatclubot

Re: In how many different ways can the letters A, A, B
[#permalink]
01 Jul 2016, 22:57

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