6 people, Adam, Bob, Charlie, Don, Eva, and Fiona have to be seated in

Question

6 people, Adam, Bob, Charlie, Don, Eva, and Fiona have to be seated in a row of six chairs numbered from 1 to 6. Adam does not want to be seated in chair numbered 1 , Bob does not want to be seated in chair numbered 2, and Charlie does not want to be seated in chair numbered 5. How many ways can they be seated so that the all the considerations of Adam, Bob, and Charlie have been taken into consideration. 
(a) 720 
(b) 714 
(c) 426 
(d) 120 
(e) None of the above

GMATBusters · Accepted Answer

We have to consider all the ways/arrangements in which Adam, Bob & Charlie do not have to compromise to seat on the chairs they don't like.
Total Number of arrangements without any restrictions = 6!= 720

We have to exclude the case when at-least one of Adam, Bob or Charlie has to seat on the chair he doesn't like.
This is infact the union of set containing the set A (when Adam has to sit on Chair 1), set B (when Bob has to sit on Chair 1),and set C (when charlie has to sit on Chair 3).
n(AUBUC)= n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C)

Now , n(A and B and C) = 3! = 6 ( as these 3 are fixed on their chair, it is simply the arrangement of remaining 3)
Similarly n(A and B) = 4!= 24, n(B and C) = 4!= 24, n(C and A) = 4!= 24
n(A) = 5! =120 ( only position of A fixed, remaining 5 can be arranged in 5!)
n(B) = 5! =120 ( only position of B fixed, remaining 5 can be arranged in 5!)
n(C) = 5! =120 ( only position of C fixed, remaining 5 can be arranged in 5!)

So, n(AUBUC) = 120+120+120-24-24-24+6 = 294
Hence the number of arrangements in which atleast one of A, B or C has to sit on chair that he doesnt like = 294

So required arrangements = total ways (without any restrictions) - number of arrangements in which atleast one of A, B or C has to sit on chair that he doesn't like.
= 6!-294 = 720-294 =426. 
Answer = C

amoghgupta · Answer

isn't the total number of combinations possible in circular arrangement=5!

amoghgupta · Answer

also, in my approach, Im using slots method, filling up the three restrictions first which gives 5*4*3 and then filling up unrestricted seats which gives 3!. So I'm getting 360 as my answer. Can someone please point out the flaw in my approach which is getting me the wrong answer?

VodkaHelps · Answer

Total number of cases is 6! = 720

Because A cannot choose 1 spot, so we have less cases by 1/6 of the total number cases. Same applies to B and C.

So the total cases approximation is: 720 x (5/6)^3 = 416 closest to 426. Answer C.

GMATGuruNY · Answer

Total arrangements = (Adam in the wrong seat) + (Bob in the wrong seat) + (Charlie in the wrong seat) - (exactly 2 in the wrong seat) - 2(all 3 in the wrong seat) + (none in the wrong seat)

Total arrangements = 6! = 720

Adam in the wrong seat: 
Number of ways to arrange the remaining 5 people = 5! = 120
 Bob in the wrong seat: 
Number of ways to arrange the remaining 5 people = 5! = 120
 Charlie in the wrong seat: 
Number of ways to arrange the remaining 5 people = 5! = 120

Exactly 2 in the wrong seat: 
 Case 1: Only Adam and Bob in the wrong seat 
Number of options for Charlie = 3 (Of the remaining 4 seats, any but seat 5)
Number of ways to arrange the remaining 3 people = 3!
To combine these options, we multiply:
3*3! = 18
 Case 2: Only Bob and Charlie in the wrong seat 
Applying the same reasoning used in Case 1 = 18
 Case 3: Only Adam and Charlie in the wrong seat 
Applying the same reasoning used in Case 1 = 18
Total ways = 18+18+18 = 54

All 3 in the wrong seat: 
Number of ways to arrange the remaining 3 people = 3! = 6

Inserting these results into the blue equation above, we get:
720 = 120 + 120 + 120 - 54 - 2(6) + (none in the wrong seat)
720 = 294 + (none in the wrong seat)
None in the wrong seat = 720 - 294 = 426

C

arvind910619 · Answer

Hello Sir,

What is wrong in the following approach.

filling up the three restrictions first which gives 5*4*3 and then filling up unrestricted seats which gives 3!. I got 360. which is clearly wrong.

GMATGuruNY · Answer

The values in red undercount the options for Adam and Bob.

Case 1: Adam chooses chair 2
Number of options for Adam = 1. (Must choose chair 2)
Number of options for Bob = 5. (Any of the 5 remaining chairs)
To combine these options, we multiply:
1*5 = 5

Case 2: Adam chooses chair 3, 4, 5 or 6
Number of options for Adam = 4. (chair 3, 4, 5 or 6)
Number of options for Bob = 4. (Of the 5 remaining chairs, any but chair 2)
To combine these options, we multiply:
4*4 = 16

Total ways = Case 1 + Case 2 = 5+16 = 21
Thus, the number of options for Adam and Bob ≠ 5*4.

Similar reasoning can be applied to the number of options for Charlie.

arvind910619 · Answer

Thanks a ton Sir,
Now I know the flaw in my reasoning; I am counting less cases.

bumpbot · Answer

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6 people, Adam, Bob, Charlie, Don, Eva, and Fiona have to be seated in

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