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# In how many different ways can the letters A, A, B

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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
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)

= (1+2+3+4+5+6+7) * 6!/(2!*3!)
= 28 * 6!/(2!*3!)
= 1680
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Re: In how many different ways can the letters A, A, B [#permalink]

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26 Jul 2014, 21:08
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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!
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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!

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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 $$\frac{3360}{2}=1680$$

Though understood the question, i could not under stand the half left and half right funda. would you please elaborate.?
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Re: In how many different ways can the letters A, A, B [#permalink]

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10 Feb 2016, 23:23
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 $$\frac{3360}{2}=1680$$

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

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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 $$\frac{3360}{2}=1680$$

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

yes it definitely helped . A big thanks. thanks B
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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

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

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