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How many different arrangements of A, B, C, D, and E are pos [#permalink]
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22 Dec 2010, 08:24
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How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D? (A) 96 (B) 60 (C) 48 (D) 36 (E) 17
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Re: Counting PS [#permalink]
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22 Dec 2010, 08:58
rxs0005 wrote: How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?
(A) 96 (B) 60 (C) 48 (D) 36 (E) 17 Total # of permutation of 5 distinct letters will be 5!=120; Glue A and B together, consider it to be one unit: {AB}{C}{D}{E} > # of permutation of these 4 units will be 4!=24, A and B within its unit also can be arranged in 2 ways : {AB} or {BA}, so total # of ways to arrange A, B, C, D, and E so that A and B to be together will be 4!*2=48; The same for A and D: total # of ways to arrange A, B, C, D, and E so that A and D to be together will be 4!*2=48; Now, the above 48+48=96 cases will contain the arrangements when A is adjacent to both B and D, so we should subtract this cases to get rid of the double counting. The # of case when A is adjacent to both B and D will be: consider {BAD} {C}{E} > # of permutation of these 3 units will be 3!, {BAD} also can be arranged in 2 ways: {BAD} or {DAB}, so total # of ways to arrange A, B, C, D, and E so that A is adjacent to both B and D will be 3!*2=12; The # of arrangements when A is adjacent to neither B nor D will be total(48+4812)=12084=36. Answer: D.
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Re: Counting PS [#permalink]
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22 Dec 2010, 09:27
rxs0005 wrote: How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?
(A) 96 (B) 60 (C) 48 (D) 36 (E) 17 Another way: as A must be adjacent to neither B nor D then it must be adjacent to only C or only E or both. Adjacent to both: {CAE}{B}{D} > # of permutation of these 3 units will be 3!, {CAE} also can be arranged in 2 ways: {CAE} or {EAC}, so total # of ways to arrange A, B, C, D, and E so that A is adjacent to both C and E will be 3!*2=12; Adjacent to only C: ACXXX (A is the first letter and C is the second): these Xs can be arranged in 3! ways. Now, it can also be XXXCA (A is the last letter and C is the fourth): again these Xs can be arranged in 3! ways. So total # of ways to arrange A, B, C, D, and E so that A is adjacent to only C is 3!*2=12; The same will be when A is adjacent to only E: 3!*2=12; Total: 12+12+12=36. Answer: D.
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Re: How many different arrangements of A, B, C, D, and E are pos [#permalink]
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01 Jul 2013, 00:59



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Re: How many different arrangements of A, B, C, D, and E are pos [#permalink]
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01 Jul 2013, 07:30
If the first letter is A, there are two options for the second letter (C or E), the remaining 3 letters can be arranged in 3! ways, thus there are 2*6=12 arrangements with A as the first letter. Similarly, if A were the last letter , there are 12 different arrangements. If A were in any one of the 3 intermediate positions, it wld have to be in between C and E  thus there would be 2 (betn C and E) *2 (betn B and D) =4 different arrangements for each of the 3 intermediate positions. Thus total number of diff arrangements = 12 (A is the first letter) + 12 (A is the last letter) + 4*3 (in any of the 3 intermediate posns) = 36.



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Re: Counting PS [#permalink]
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21 May 2014, 06:33
Bunuel wrote: rxs0005 wrote: How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?
(A) 96 (B) 60 (C) 48 (D) 36 (E) 17 Total # of permutation of 5 distinct letters will be 5!=120; Glue A and B together, consider it to be one unit: {AB}{C}{D}{E} > # of permutation of these 4 units will be 4!=24, A and B within its unit also can be arranged in 2 ways : {AB} or {BA}, so total # of ways to arrange A, B, C, D, and E so that A and B to be together will be 4!*2=48; The same for A and D: total # of ways to arrange A, B, C, D, and E so that A and D to be together will be 4!*2=48; Now, the above 48+48=96 cases will contain the arrangements when A is adjacent to both B and D, so we should subtract this cases to get rid of the double counting. The # of case when A is adjacent to both B and D will be: consider {BAD} {C}{E} > # of permutation of these 3 units will be 3!, {BAD} also can be arranged in 2 ways: {BAD} or {DAB}, so total # of ways to arrange A, B, C, D, and E so that A is adjacent to both B and D will be 3!*2=12; The # of arrangements when A is adjacent to neither B nor D will be total(48+4812)=12084=36. Answer: D. Looks good only thing I got wrong was that on the last step namely: 'Now, the above 48+48=96 cases will contain the arrangements when A is adjacent to both B and D, so we should subtract this cases to get rid of the double counting. The # of case when A is adjacent to both B and D will be: consider {BAD} {C}{E} > # of permutation of these 3 units will be 3!, {BAD} also can be arranged in 2 ways: {BAD} or {DAB}, so total # of ways to arrange A, B, C, D, and E so that A is adjacent to both B and D will be 3!*2=12;' I subtracted 12 twice that is 24, because BAD and DAB are included in both the first case with AB together and the second case with AD together Could you please explain why you only subtract once and not twice, I've made this error several times already and I can't seem to get the grip on this issue Thanks! Cheers J



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Re: Counting PS [#permalink]
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21 May 2014, 06:57
jlgdr wrote: Bunuel wrote: rxs0005 wrote: How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?
(A) 96 (B) 60 (C) 48 (D) 36 (E) 17 Total # of permutation of 5 distinct letters will be 5!=120; Glue A and B together, consider it to be one unit: {AB}{C}{D}{E} > # of permutation of these 4 units will be 4!=24, A and B within its unit also can be arranged in 2 ways : {AB} or {BA}, so total # of ways to arrange A, B, C, D, and E so that A and B to be together will be 4!*2=48; The same for A and D: total # of ways to arrange A, B, C, D, and E so that A and D to be together will be 4!*2=48; Now, the above 48+48=96 cases will contain the arrangements when A is adjacent to both B and D, so we should subtract this cases to get rid of the double counting. The # of case when A is adjacent to both B and D will be: consider {BAD} {C}{E} > # of permutation of these 3 units will be 3!, {BAD} also can be arranged in 2 ways: {BAD} or {DAB}, so total # of ways to arrange A, B, C, D, and E so that A is adjacent to both B and D will be 3!*2=12; The # of arrangements when A is adjacent to neither B nor D will be total(48+4812)=12084=36. Answer: D. Looks good only thing I got wrong was that on the last step namely: 'Now, the above 48+48=96 cases will contain the arrangements when A is adjacent to both B and D, so we should subtract this cases to get rid of the double counting. The # of case when A is adjacent to both B and D will be: consider {BAD} {C}{E} > # of permutation of these 3 units will be 3!, {BAD} also can be arranged in 2 ways: {BAD} or {DAB}, so total # of ways to arrange A, B, C, D, and E so that A is adjacent to both B and D will be 3!*2=12;' I subtracted 12 twice that is 24, because BAD and DAB are included in both the first case with AB together and the second case with AD together Could you please explain why you only subtract once and not twice, I've made this error several times already and I can't seem to get the grip on this issue Thanks! Cheers J (a) The cases for which A and B are together (48) include the cases cases when A is adjacent to both B and C: {the cases when A is adjacent only to B} + {the cases when A is adjacent to both B and C}. (b) The cases for which A and C are together (48) include the cases cases when A is adjacent to both B and C: {the cases when A is adjacent only to C} + {the cases when A is adjacent to both B and C}. (c) The number of cases when A is adjacent to both B and C is 12. Now, to get the number of cases for which A is adjacent to B, or C or both = {the cases when A is adjacent only to B} + {the cases when A is adjacent only to C} + {the cases when A is adjacent to both B and C}, which is (a) + (b)  (c). Basically the same way as when we do for overlapping sets when we subtract {both}: {total} = {group 1} + {group 2}  {both}. Does this make sense?
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Re: Counting PS [#permalink]
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21 May 2014, 08:16
Oh ok, gotcha. Yeah the thing is that when in overlapping sets you only want to count the members of Set A or B, but not both then it is correct to subtract 'Both' two times. Say like How many of the multiples of 3 and 5 are not multiples of 15? Then you would only take the multiples of 3 and 5 and subtract 2* (Multiples of 15). This reasoning doesn't quite apply to this question as we do in fact want to consider the scenario in which all three are seated together. Therefore, we should use {both}: {total} = {group 1} + {group 2}  {both} as you correctly mentioned Clear now Thanks Cheers! J



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How many different arrangements of A, B, C, D, and E are pos [#permalink]
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03 Jun 2015, 04:49
rxs0005 wrote: How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?
(A) 96 (B) 60 (C) 48 (D) 36 (E) 17 ALTERNATE METHOD:We can make cases here Case 1: A takes position no.1 i.e. Arrangement looks like (A _ _ _ _)In this case B and D can take any two position out of position no.3, 4, and 5 i.e. B and D can take position in 3x2 = 6 ways remaining two letters C and E can be arranged on remaining two places in 2! ways = 2 ways i.e. total arrangement as per Case 1 = 6 x 2 = 12 ways Case 2: A takes position no.2 i.e. Arrangement looks like (_ A _ _ _)In this case B and D can take any two position out of position no. 4 and 5 i.e. B and D can take position in 2! = 2 ways remaining two letters C and E can be arranged on remaining two places in 2! ways = 2 ways i.e. total arrangement as per Case 2 = 2 x 2 = 4 ways Case 3: A takes position no.3 i.e. Arrangement looks like (_ _ A _ _)In this case B and D can take any two position out of position no. 1 and 5 i.e. B and D can take position in 2! = 2 ways remaining two letters C and E can be arranged on remaining two places in 2! ways = 2 ways i.e. total arrangement as per Case 3 = 2 x 2 = 4 ways Case 4: A takes position no.4 i.e. Arrangement looks like (_ _ _ A _) This case is same as Case 2 (Just mirror of case 2) hence total ways of arrangement will remain 4 ways onlyIn this case B and D can take any two position out of position no. 1 and 2 i.e. B and D can take position in 2! = 2 ways remaining two letters C and E can be arranged on remaining two places in 2! ways = 2 ways i.e. total arrangement as per Case 4 = 2 x 2 = 4 ways Case 5: A takes position no.5 i.e. Arrangement looks like (_ _ _ _ A)This case is same as Case 2 (Just mirror of case 1) hence total ways of arrangement will remain 4 ways onlyIn this case B and D can take any two position out of position no.1, 2, and 3 i.e. B and D can take position in 3x2 = 6 ways remaining two letters C and E can be arranged on remaining two places in 2! ways = 2 ways i.e. total arrangement as per Case 5 = 6 x 2 = 12 ways Total Ways of favorable arrangements = 12+4+4+4+12 = 36 ways Answer: Option
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How many different arrangements of A, B, C, D, and E are pos [#permalink]
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22 Oct 2016, 07:36
A will be adjacent to B or D, in any of the permutations of ABD.
So total permutations of "CE[ABD]" is 3! * 3!
Total permutations of "ABCDE" is 5!.
Shouldn't the answer be 5!  3! * 3!. What is it that I'm missing ?



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Re: How many different arrangements of A, B, C, D, and E are pos [#permalink]
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17 Nov 2016, 21:11
I'm a visual person, so maybe my explanation will help those who are having difficulty seeing what's going on in this problem: Given that we have 5 different letters to order, there are 5 places A could be situated  thus, we will devise 5 cases
(1) A __ __ __ __ (2) __ A __ __ __ (3) __ __ A __ __ (4) __ __ __ A __ (5) __ __ __ __ A
We also know that B or D cannot be adjacent to A, so we will adjust our scenarios to reflect such:
(1) A (~B/D) __ __ __ (2) (~B/D) A (~B/D) __ __ (3) __ (~B/D) A (~B/D) __ (4) __ __ (~B/D) A (~B/D) (5) __ __ __ (~B/D) A
Now, looking at the scenarios, starting with (1), we see that position two will have to be taken by either C or E. This will give us two baseline scenarios to work off of
(a) A C __ __ __ (b) A E __ __ __
Now the remaining three slots available can be occupied by a combination of the remaining 3 letters > 3! Thus, for each scenario within main scenario (1) (i.e. A in first slot), we will have 12 combinations (i.e. 6 each)
This will also be the case for (5), since A will be basically flipped to the opposite side > so we can say right now that (5) will also have 12 combinations
(2) C/E A C/E __ __ > breaks down into two scenarios again
(a) C A E __ __ > two combinations (with remaining letters) (b) E A C __ __ > two combinations (with remaining letters)
Thus (2) gives us 4 combinations
(4) will also give us 4 combinations for the same reason
(3) __ C/E A C/E __ remains > break into scenarios once more
(a) __ C A E __ > 2 combinations possible (with remaining letters) (b) __ E A C __ > 2 combinations possible (with remaining letters)
Thus, adding (1)+(2)+(3)+(4)+(5) = 36 combinations



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Re: How many different arrangements of A, B, C, D, and E are pos [#permalink]
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18 Nov 2016, 04:33
rxs0005 wrote: How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?
(A) 96 (B) 60 (C) 48 (D) 36 (E) 17 If A can be adjacent to neither B nor D, it would be either nestled between E and C or at an extreme position next to E or next to C. No of cases with A nestled between E and C = 3! * 2! = 12 No of cases with A at left extreme with C/E next to it = 2*3! = 12 (Choose one of C or E in 2 ways and arrange the remaining 3 letters) No of cases with A at right extreme with C/E next to it = 12 (same as above) Total = 12+12+12 = 36 Answer (D)
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How many different arrangements of A, B, C, D, and E are pos [#permalink]
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28 Mar 2018, 23:22
rxs0005 wrote: How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?
(A) 96 (B) 60 (C) 48 (D) 36 (E) 17 Leftmost arrangement of constraint elements is A_BD_ Using formula we have (2!*2)*3*2!=24 B_A_D Using formula we have 2!*3!*1=12 The total number of permutations is 36 For explanation of this formula and other types of problems kindly visit the link below.
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