2 American men; 2 British men; 2 Chinese men and one each of Dutch, Eg

Question

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

A. 30
B. 45
C. 60
D. 64
E. 72

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ScottTargetTestPrep · Accepted Answer

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   and  , respectively. Thus, we are arranging  ,  , 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  ,  ,  , 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,  ,  , D, E, F, G or  , B, b,  , E, F, G in addition to  ,  , 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

CrackverbalGMAT · Answer

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)

Bunuel · Answer

___________________
Fixed. Thank you!

abhik1502 · Answer

Bunuel

will you pls share the OA.

I solved and got 72 as the answer.

My approach :
2 pair will sit together. So if we count 2 pair as 1 unit then on round table we need to arrange

4 (one each of Dutch, Egyptian, French and German) + 2 (a pair that will not sit together with other 2) + 1 (2 pairs that will sit together and that has been counted as 1 unit till now) = 7

So we can arrange above 7 units in 6! ways.

Now, Out of 3 pairs 2 need to be selected for sitting together = 3C2
We can arrange above 2 pair of people in 4! ways.

So total arrangement = 3C2 x 4! x 6! = 72 x 6!

41396302717 · Answer

Hello ScottTargetTestPrep

I went ahead with this approach:

Number of ways to select 2 couples out of 3 is 3C2 and then each couple can be rotated therefore selection of 2 couples and their internal rotation sums to - 3C2 * 2! * 2!

After we have 2 couples we are left with 8 places to fill up the circular permutation which gives us the permutation 7! but this permutation will also have those permutations in which all 3 couples are together.

Now we need to subtract those permutations : we can get those permutation as follows 3C3 * 6! * 2! * 2! * 2!.

Now all that's left to do is subtract these terms

therefore the answer comes out to be: (7! * 3C2 * 2! * 2!) - (6! * 2! * 2! * 2!) => (84 * 6!) - (8 * 6!) => 76 * 6!

Please help me find the flaw in this reasoning.

Thank you in advance!

ScottTargetTestPrep · Answer

Response:
 
In short, your mistake is that you are overcounting the arrangements where all three couples are together, but you are subtracting that number only once from 84 * 6!.

To better illustrate your mistake, let’s denote the couples by A, B and C. There are 3C2 ways to pick two couples from three, one of those ways is  . One of the arrangements you included in 84 * 6! possible arrangements is  . On the other hand,   is also one of the ways included in the 3C2 ways to pick two couples. Thus,   is an arrangement that you included in 84 * 6! possible arrangements. However, since this is a circular permutation problem, the arrangements   and   are actually the same arrangement. Moreover,   is also the same arrangement as the previous two. You counted all these arrangements as separate possibilities, but you subtracted only one of these arrangements from 84 * 6!. If you subtract three times the number you calculate to be the number where three couples are sitting together, you will obtain 84 * 6! - 3 * 8 * 6! = 60 * 6!, and thus p = 60.

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2 American men; 2 British men; 2 Chinese men and one each of Dutch, Eg

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