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 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

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

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coolwhizRaj · Accepted Answer

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

TestPrepUnlimited · Answer

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

Crytiocanalyst · Answer

The key to solving the question is getting all the equation right 
American couple is adjacent = 8!2!
where British adjacent - 7!2!2!
where Chinese adjacent-7!2!2!
both British and Chinese adjacent- 6!2!2!2!
ONly the americans are allowed to sit together others are not allowed to sit together that's the chinese and brits 
=> 8!2! - 7!2!2! - 7!2!2! + 6!2!2!2!
=>6! (56*2 - 28 - 28 + 8)
=>6! 64
Hence IMO D

VirFortis · Answer

American(A), British(B) and Chinese(C)

Step 1 : If both A together -> treat them as single entity -> So, arrangement to be done for 9 entities in a circle -> (n-1)! * arrangement within the single entity of As -> 8!*2! = Total arrangments.
Step 2: We need to remove cases of Bs adjacent and Cs adjacent ->
 a. Bs adjacent -> treat As together as an entity, treat Bs as an entity -> So, arrangement to be done for 1+1+6 entities -> 7!*2!*2!
 b. Cs adjacent -> treat As together as an entity, treat Cs as an entity -> So, arrangement to be done for 1+1+6 entities -> 7!*2!*2!
Step 3: While removing the adjacent cases in Step 2, we also removed scenarios when all three A, B and C pairs are together, but we removed it twice - in 2a and 2b. But we need to remove that case only once. Hence, we just add that scenario.
As together, Bs together, Cs together -> arrangement for 3+4 -> 6!*2!*2!*2!

Step 4: Calculate from Step1 to Step3 -> 8!2! - 7!2!2! - 7!2!2! + 6!2!2!2! = 6!(56*2 - 28 - 28 + 8) 
= 6!(56(2-1)+8)=6!(64)
Answer : D.

PS: I got this wrong, but tried to analyse and understand the approach for this. Posting the explanation, so that this helps others in analysis.

SHEHAB01 · Answer

1.Why you consider only british and chinese persons else other dutch, egyptian, french and german persons?

2.I can't find out the approaches 8!2!,7!2!2!, and 6!2!2!2!.I do it by like(8!/2!)...can't really understand 😭😭Help please

Posted from my mobile device

bumpbot · Answer

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

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