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19 Feb 2021, 03:15
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C
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% (02:48) correct
75%
(02:55)
wrong
based on 102
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 they can be seated if exactly two pairs of persons of same nationality are together is p*(6!), then what is the value of p ?
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
09 Mar 2021, 14:23
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Bunuel
Solution:
Let’s first calculate the number of ways in which the two Americans and the two British are sitting together, but the two Chinese are not. We will first calculate the number of arrangements without the condition that the Chinese people are not sitting together. Next, we will calculate the number of arrangements where the Chinese people ARE sitting together. We will then easily obtain the number of arrangements where the Chinese people ARE NOT sitting together by subtracting the latter from the former.
Since the two Americans and the two British are sitting together, we can consider each as a single entity, which we will denote by [Aa] and [Bb], respectively. Thus, we are arranging [Aa], [Bb], C, c, D, E, F, G around a circular table.
Since there are 8 items and since this is circular permutation, there are (8 - 1)! = 7! ways of arranging the 8 items. However, notice that the two Americans can sit together either as Aa or as aA. Similarly, the British can sit together either as Bb or as bB. Thus, there are 7! x 2 x 2 = 4*7! ways where the Americans and the British are sitting together.
Now, let’s calculate the number of ways where the Chinese are also sitting together in addition to the Americans and the British. Similar to the above case, we need to arrange [Aa], [Bb], [Cc], D, E, F, G around a circular table. Now, we have 7 items and there are (7 - 1)! = 6! ways to arrange 7 items around a circle. In this scenario, in addition to the Americans and the British, the Chinese can also sit together in two ways: as Cc or as cC. Thus, there are 6! x 2 x 2 x 2 = 8*6! ways where the American, the British and the Chinese are sitting together.
Subtracting the number of arrangements where the Chinese are sitting together from the number of arrangements where only Americans and British are sitting together, we see that there are 4*7! - 8*6! = 4*6!(7 - 2) = 4*6!*5 = 20*6! ways where the Chinese are not sitting together.
We calculated the number of ways where the American and the British are sitting together but the Chinese are not; however, that is not the only way of having the condition “exactly two pairs of people from the same nationality are sitting together. We could have A, a, [Bb], [Cc], D, E, F, G or [Aa], B, b, [Cc], E, F, G in addition to [Aa], [Bb], C, c, D, E, F, G. In each of these cases, there are 20 * 6! ways to seat the people around a circular table. Thus, there are 20*6! + 20*6! + 20*6! = 60*6! ways of seating the people around the circular table satisfying the requirements of the question.
Solving 60*6! = p*6!; we obtain p = 60.
Answer: C
General Discussion
19 Feb 2021, 05:10
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Bunuel Hi! If you could edit the last line of the question, it would turn clear and easier to solve.
Thank You
Devmitra Sen(Math)
Thank You
Devmitra Sen(Math)
