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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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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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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.
Re: In how many different ways can the letters A, A, B [#permalink]
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08 Nov 2016, 13:57
Hey i would like to know if the answer will change if the question is : In how many different ways can the letters A, A, B, B, C, D, E,E be arranged if looking at permutations of 8 letter words consisting of the above, where A doesnt appear before B
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