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

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Intern
Joined: 15 Feb 2013
Posts: 5
Re: In how many different ways can the letters A, A, B  [#permalink]

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26 Jun 2013, 14:10
1
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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Posts: 101
Re: In how many different ways can the letters A, A, B  [#permalink]

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23 Sep 2014, 10:00
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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01 Jul 2016, 21:57
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
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Posts: 8296
Re: In how many different ways can the letters A, A, B, B,  [#permalink]

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11 Feb 2018, 17:55
henrymba2021 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. 1,680
b. 2,160
c. 2,520
d. 3,240
e. 3,360

Hi..
First let's calculate total ways..
What do we have ..
2*A, 3*B and 1 of C,D and E..
So total 8..
Ways of combination= 8!/2!3!=8*7*5*4*3=3360..
But in these half will have C on right of D and half D on right of C..
So ans = 3360/2=1680

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

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11 Feb 2018, 19:50
1
henrymba2021 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. 1,680
b. 2,160
c. 2,520
d. 3,240
e. 3,360

These are the arrangements possible where letter C is to the right of D
D******C
D*****C*
D****C**
D***C***
D**C****
D*C*****
DC******(7 combinations when D is the first alphabet)
*D*****C
*D****C*
*D***C**
*D**C***
*D*C****
*DC*****(6 combinations when D is the second alphabet)
**D****C
**D***C*
**D**C**
**D*C***
**DC****(5 combinations when D is the third alphabet)
The total arrangements possible are $$7+6+5+4+3+2+1 = 28$$

The total ways in which alphabets A,A,B,B,B,C can be arranged are$$\frac{6!}{2!*3!} = \frac{6*5*4*3*2}{2*3*2} = 6*5*2 = 60$$

Therefore, the total ways in which the alphabets can be arranged when C is to the right of D is 28*60 = 1680(Option A)
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Re: In how many different ways can the letters A, A, B  [#permalink]

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03 Mar 2018, 02:10
Can someone help me with this problem . really not able to understand the logic?
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Re: In how many different ways can the letters A, A, B  [#permalink]

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03 Mar 2018, 02:23
Intern
Joined: 11 Feb 2018
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Re: In how many different ways can the letters A, A, B  [#permalink]

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30 Jan 2020, 03:45
The question says C has to be to the right of D not necessarily in consequtive places. So if D takes the first position then C has 7 positions as there are 8 places. Then if D takes 2nd place there are 6 places to it's right for C. If D is in 3rd position than C has 5 and so on and the last position D can take is the 7th place so that C can take the 8th place. So in all D with C to it's right is possible in 7+6+5+4+3+2+1=28 positions. For each of these, the rest 6 can arrange themselves in 6!. Since there are repetitions, the total ways for the rest 6 of them will be 6!/(2!*3!)
So the total arrangements with C to the right of D will be 28*6!/(2!*3!)= 1680. So A

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Re: In how many different ways can the letters A, A, B   [#permalink] 30 Jan 2020, 03:45

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