Question

A, T, J, K, and C, win tickets to a Broadway show. The five seats are all together in one row of the theater. i) How many ways are there for the five friends to sit in the seats? ii) If they each randomly pick one ticket, what is the probability that A and T will sit together?

Answer 1

(i)

Number of ways are there for the five friends to sit in the seats = 5! = 5 X 4 X 3 X 2 X 1 = 120

So,

Answer is:

120

(ii)

Since A and T will sit together, consider AT as one item.

Thus,

AT, J, K and C can be permuted in 4! = 4 X 3 X 2 X 1 = 24

Since A and T can be permuted in 2 ways, number of ways of A and T to be together = 24 X 2 = 48

So,

the probability that A and T will sit together = 48/120

                                                              = 0.4

So,

Answer is:

0.4