I did the first question like selecting 2 seats from 20 seats 20c2 and subtracting the no. of ways in which two adjacent pairs can selected like12,23,34,…etc in this way i m getting 20c2 -19 =171

In the first question, I just wanted to be certain as to why you have divided by two.
Is it because we cannot distinguish between the two ends, i.e., a person sitting towards the end is one possibility (and we are not talking about right end or left end in this question).

So, if the question had seat there are 20 seats “numbered” 1 to 20, then we would have divided by 2, right?

No, it is because the order of seats chosen is not important. If I started with seat number 5 and then seat number 9, or if I started with seat number 9 and then picked 5 to go with it, it would be counted as two separate cases in the 342. But actually it is the same case as we are just selecting two seats.

For the second question,let’s say we choose 2 seats a and b.Let’s take there are t1 seats before a,t2 seats between a and b and t3 seats after b.Minimum value for t2 is 3.So t1+t2+t3=15.So total 17C2=136 no. of ways are possible.

I didn’t understand the solutions.

Hi J,

Can you please explain Approach #2? How will the number of available seats be 16 if Seat 3 is chosen as first one?

Seat numbers 5 to 20 = 16 seats.

regards

J

I did the first question like selecting 2 seats from 20 seats 20c2 and subtracting the no. of ways in which two adjacent pairs can selected like12,23,34,…etc in this way i m getting 20c2 -19 =171

Yes, that’s a quick and perfectly valid approach as long as there are only two seats to be chosen ðŸ™‚ Becomes messy for more than 2 of course….

regards

J

In the first question, I just wanted to be certain as to why you have divided by two.

Is it because we cannot distinguish between the two ends, i.e., a person sitting towards the end is one possibility (and we are not talking about right end or left end in this question).

So, if the question had seat there are 20 seats “numbered” 1 to 20, then we would have divided by 2, right?

No, it is because the order of seats chosen is not important. If I started with seat number 5 and then seat number 9, or if I started with seat number 9 and then picked 5 to go with it, it would be counted as two separate cases in the 342. But actually it is the same case as we are just selecting two seats.

regards

J

For the second question,let’s say we choose 2 seats a and b.Let’s take there are t1 seats before a,t2 seats between a and b and t3 seats after b.Minimum value for t2 is 3.So t1+t2+t3=15.So total 17C2=136 no. of ways are possible.