## SIG (Susquehanna International Group) Interview Question

Analysts**Country:**United States

Ty for the answer. I was wondering why you have not considered that they might be sitting in anti clockwise direction as well ? The above answer will apply even if we speak of a straight table. But its a round table...Can anyone help here..? My answer is also coming 1/12 using addition theory. :/

Total no of combinations in a round table arrangement = (n-1)!

Ordering we are interested = 2 (PS. Ascending and descending order are different as for a person, the one sitting on left and the one sitting on right will be interchanged when the ordering moves from ascending to descending)

So, probability = 2/(n-1)! = 2/4! = 1/12

assuming the cont happens only in one direction either clockwise or anti.

the first person with least age can select any of the 5 available seats.next person has only one way ans so on upto 5th person.

Same for descending there are 5 ways.

so total is 10 ways

the no of total ways is 5! as we are counting only in one direction.

so probability is 10/5!

Very easy questions.

We fix a place where the person with the lowest age sits.

Now, the probability that the person with the second lowest age sits either to his left = (number of seats he can sit on / total number of seats left) = 1/4.

The probability that the person with the third lowest age sits two seats left of the person with the lowest age = 1/3 since there are 3 seats left.

Similarly, probability that next person sits in the right place is 1/2.

After this, the 5th person only has one seat to sit on.

So the probability of all these events happening together = 1/2 * 1/3 * 1/4 = 1/24.

Now the same arrangement could happen towards the right (opposite direction). Either of the arrangements is possible.

So total probability = 1/24 * 2 = 1/12.

Let the people be A, B, C, D, E in order of age.

A can sit anywhere, so he sits "correctly" with probability 1.

B has to sit to A's immediate left or right for things to work, so this happens with probability 2/4 = 1/2 (of 4 remaining seats)

C has to sit next to B, with probability 1/3 (of 3 remaining seats).

D has to sit next to C, with probability 1/2 (of 2 remaining seats)

E sits in the last remaining spot with probability 1.

Therefore the total probability is 1*1/2*1/3*1/2*1 = 1/12.

the simplest of all answers

(n-1)! ways off arrngng n persons in a round table. here it is 4!.

since here there is no diffrence between clockwise order n anticlockwise order, so arrangement would be (n-1)!/2 , i.e 4!/2 = 12.

there is only one way to arrange all people in ascending order(clockwise) which is also a descending order(anticlockwise)

so the probabilty is 1/12.

Imagine that each seat around the round table is numbered 1,2,..,5

- brighama March 25, 2013Imagine that each person is numbered 1,2,...,5 according to their age-rank.

There are 5 ways that the 5 people can be seated in ascending-age order.

There are 5 ways that the 5 people can be seated in descending-age order.

There are 5! different ways the people can be seated in general

So, the probability of a successful arrangement is (5+5)/5! or 1/12