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12 Feb 2021, 03:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
25% (02:49) correct
75%
(02:42)
wrong
based on 113
sessions
History
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Time
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Not Attempted Yet
2 American men; 2 British men; 2 Chinese men and one each of Dutch, Egyptian, French and German persons are to be seated for a round table conference. If the number of ways in which only American pair is adjacent is equal to q*(6!), then what is the value of q ?
A. 30
B. 45
C. 60
D. 64
E. 72
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
A. 30
B. 45
C. 60
D. 64
E. 72
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
12 Jul 2021, 02:32
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Solution = instances where American couple is adjacent - where British adjacent - where Chinese adjacent + both British and Chinese adjacent
= 8!2! - 7!2!2! - 7!2!2! + 6!2!2!2!
=6! (56*2 - 28 - 28 + 8)
=6! 64
= 8!2! - 7!2!2! - 7!2!2! + 6!2!2!2!
=6! (56*2 - 28 - 28 + 8)
=6! 64
General Discussion
13 Feb 2021, 20:46
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Bunuel
There is probably a better solution than mine but I will approach this question by expanding a table into 10 slots and use a slotting method. The trick is at the end we divide by 10 because there are 10 options that form the same round table when there are 10 seats.
First, treat A1 and A2 as one unit (but at the end we have to multiply by 2 since we can swap the A's to form a new table). Then we have only 9 units, to begin with, there are 9! options using the slot method with the Americans paired, but we need to add the options where the A's are at the beginning and the end of the slots, as when we form the round table that results in the A's being together. Thus with those fixed, we have 8 slots in the middle, giving us 8! more options. Then we have 9! + 8! options or 8! * 10 with the A's paired up.
Next, we want to eliminate the options with (1) B's paired, (2) C's paired. Eliminating both (1) and (2) results in the options with both pairs B and C being eliminated twice, so we want to add back the options with (3) B and C both paired.
Thus the answer would look like this: 2 * (8! * 10 - (1) - (2) + (3)) / 10.
(1) and (2) conveniently have the same value. This is under the circumstance of the A's treated as one unit so we have only 9 slots now. Combining B's or C's gives us only 8 slots. Thus we start off with 8! selections, but again we need to consider the pair split up at the first and last slot. Using the pattern from above it is 7! options from that scenario. Also, don't forget the scenario with A's on the endpoints, that's another 7! options so we have 8! + 7!*2 options total. We need to multiply by 2 since we can swap the pairs' position.
Thus (1) = (2) = (8! + 7! * 2) * 2, and (1) + (2) = 8! * 4 + 7! * 8
(3) happens if both B and C are paired. Now we have only 7 slots with the pairs B and C treated as one unit, so 7! options. For one last time consider either B or C split up at the endpoints, and the other pair in between them, that would be 6! options. We get another 6! when we also consider the scenario with A's on the endpoints, therefore 7! + 6! * 2 options. We can swap B's or C's so multiply that by 4.
Thus (3) = (7! + 6! *2)*4 = 7! * 4 + 6! * 8
Finally, 2 * (8! * 10 - (1) - (2) + (3)) = \(2*(8!*10 - 8! * 4 - 7! * 8 + 7! * 4 + 6! * 8) = 2*(8!*6 - 7!*4 + 6!*8)\)
Divide by 6! to get \(2*(8*7*6 - 7*4 + 8) = 2*340 = 2*(7*44 +8) = 2*(312+8) = 2*320 = 640\).
Finally divide by 10 to get 64.
Ans: D
