Question

Problem 5.There are thirty people attending a show in an amphitheater which has six rows ofseating. Assume that it is possible (but not necessary) for everyone to sit in a single row, and thatwe don’t care about the amount of space between people in the same row.

(a) How many ways may these thirty people sit in the amphitheater?

(b) How many ways may these thirty people sit in the amphitheater, if no row is to be leftempty?

(c) How many ways may these thirty people sit in the amphitheater, if Anne and Bob must sitnext to each other?

(d) How many ways may these thirty people sit in the amphitheater, if there must be exactlyfour people in the first row and at least one person in the last row?

Answer 1

Let x1, x2, x3, x4, x5 and x6 be the number of peope in row 1 to row 6 respectively

(a) The number of ways is the number of solutions of the equation

x1 + x2 + x3 + x4 + x5 + x6 = 30, xi ≥ 0

which is equal to = 35C5 = 324,632

(b) The number of ways in this case is the number of solutions of the equation x1 + x2 + x3 + x4 + x5 + x6 = 30 , xi ≥ 1

which is equal to = 29C5 = 118,755

(c) If Anne and Bob must sit next to each other, the remaining 28 people need to be seated in the six rows.

Anne and Bob can sit in any of the 6 rows

The number of ways would be (the number of solutions of the equation

x1 + x2 + x3 + x4 + x5 + x6 = 28, xi ≥ 1)*6

= *6 = *6 = 1,424,016

(d) Four people are assigned in row 1 and 1 person is assigned in row 6.

Thus, the remaining 25 people can sit in any of the row 2 to row 6.

The number of ways = = 29C4 = 23,751