sunita123 wrote:
Hi,
Can i solve this by taking total number of arrangements i.e 5!=120 ways and then subtract the restriction?
120-(arrangements B should not sit next to A and/or C) ?
is that a correct approach?
GMATinsight wrote:
Quote:
In how many ways can 6 people A,B,C,D,E,F be seated at a round table if B cannot sit next to A and/or C?
a) 720
b) 120
c) 108
d) 84
e) 36
Have Modified the Language to make it clearer6 people are A, B, C, D, E and F
and B can not sit next to A and C
Considering the Position of B is fixed,
We have to make sure that 2 person who sit next to B are out of D, E and F
i.e. No. of ways of choosing the neighbours of B = 3
C2 = 3
and the no. of ways the Selected neighbours can arrange at the two position adjacent to B = 2!
i.e. The ways the B and The neighbours can be arranged = 3
C2 *2! = 3*2 = 6
Now The no. of ways in which Remaining Three Individuals can be arranged on remaining 3 seats = 3!
Total Ways of making Six person seated such that B doesn't sit next to A and C = 3
C2 *2!*3! = 36
That would be fine but a difficult approach as you will have to calculate three casesCase-1: When A sits next to B and C does not sit next to B
A can sit next to B in 2 ways (On B's right or B's left side)
The next adjacent place of B can be occupied in 3 ways because C can't sit next to B
Remaining three can sit in 3! ways
So total ways = 2*3*3! = 36 ways
Case-2: When C sits next to B and A does not sit next to B
C can sit next to B in 2 ways (On B's right or B's left side)
The next adjacent place of B can be occupied in 3 ways because A can't sit next to B
Remaining three can sit in 3! ways
So total ways = 2*3*3! = 36 ways
Case-3: When A and C both sit on either sides of B
A and C can sit in 2! ways on two places adjacent to B
Remaining three can sit in 3! ways
So total ways = 2!*3! = 12 ways
Total Unfavourable cases = 36+36+12 = 84 ways
Total favourable Cases = (6-1)! - 84 = 120 - 84 = 36 ways
I hope it helps!
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