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5 friends, A, B, C, D and E, sit next to each other in 5 adjacent seat
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16 Feb 2020, 03:46
Question Stats:
48% (01:30) correct 53% (01:29) wrong based on 40 sessions
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5 friends, A, B, C, D and E, sit next to each other in 5 adjacent seats in a row. How many different arrangements are possible when exactly one person is between A and C? a. 12 b. 18 c. 36 d. 72 e. 120
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Re: 5 friends, A, B, C, D and E, sit next to each other in 5 adjacent seat
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16 Feb 2020, 07:52
First, let's put A to the left of C. Then we know somewhere in our seating arrangement, we have AXC, where X is a person sitting between A and C. In this case, there are only three choices for where to put person A  the first, second, or third seats. We can't put A in the fourth or fifth seat, because there's no sixth or seventh seat where we could put person C. So in this case, we have 3 choices for where to put A, after which the position of C is fixed, and then we still have 3 empty seats, and we can seat the remaining people in those seats in (3)(2)(1) = 6 ways. So, multiplying our choices, we have (3)(6) = 18 seating arrangements in total. But we could also put C to the left of A. We'll again find we have 18 possible seating arrangements. So in total, there are 18+18 = 36 arrangements.
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5 friends, A, B, C, D and E, sit next to each other in 5 adjacent seat
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16 Feb 2020, 08:08
When A and C are on seats 1 and 3 respectively, we have 3 options to choose the person who sits between A and C and the remaining two people can arrange themselves in 2!
So when A and C are on seats 1 and 3 respectively, we have 3*2!=6 arrangements
Similarly when A and C sit on seats 2 and 4, we have 6 arrangements
Similarly when A and C sit on seats 3 and 5, we have 6 arrangements
Now the same repeats when A and C switch positions in each of the above cases
So total number of arrangements is 6+6+6+6+6+6=36
Answer is (C)
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Joined: 11 Feb 2017
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Re: 5 friends, A, B, C, D and E, sit next to each other in 5 adjacent seat
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28 May 2020, 12:22
1. There is 3C1 ways of selecting the one person between A and C 2. 2 ways of selecting A first or C. i.e. AXC or CXA 3. Suppose AXC to be one, then there are three friends, namely: 'AXC', 'remaining two'. These three can be arranged in 3! ways.
Therefore Total Arrangements: 3C1*2*3! = 36



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Re: 5 friends, A, B, C, D and E, sit next to each other in 5 adjacent seat
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28 May 2020, 13:30
C. 36 Let's assume: Scenario 1: A _ C _ _ => Ways to fill remaining places = 3*2*1 = 6; A & C can also switch in 2 ways resulting is different seating arrangement = 6 *2 =12 Scenario 2: _ A _ C _ => Ways to fill remaining places = 3*2*1 = 6; A & C can also switch in 2 ways resulting is different seating arrangement = 6 *2 =12 Scenario 3: _ _ A _ C => Ways to fill remaining places = 3*2*1 = 6; A & C can also switch in 2 ways resulting is different seating arrangement = 6 *2 =12 Total: 12 + 12 + 12 =36
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Re: 5 friends, A, B, C, D and E, sit next to each other in 5 adjacent seat
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28 May 2020, 13:30




