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Re: In how many different ways can the letters A, A, B
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26 Jun 2013, 15:10
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



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Re: In how many different ways can the letters A, A, B
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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
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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



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Re: In how many different ways can the letters A, A, B, B,
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11 Feb 2018, 18:55
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,
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11 Feb 2018, 20:50
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
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03 Mar 2018, 03:10
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
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Re: In how many different ways can the letters A, A, B
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18 Mar 2019, 17:46
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Re: In how many different ways can the letters A, A, B
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