6 people consists of 3 married couples. Each couple wants to sit with older partner on the left. Required:
a. How many ways can they be seated together in the row?
b. Suppose one of the six is a doctor who must sit on the aisle in case she is paged. How many ways can the people be seated together in the row with the doctor in an aisle seat?
c. Suppose the six people consist of three married couples and each couple wants to sit together with the husband on the left. How many ways can the six be seated together in the row?

a. The first part asks for how many ways they can be seated together in a row. Therefore we want the permutations of the set of 6 people, or 6 factorial,

6! = 6 5

= 30 4

= 360 2 = 720 possible ways to order 6 people in a row

b. There are two cases to consider here. If the doctor were to sit in the left - most seat, or the right - most seat. In either case there would be 5 people remaining, and hence 5! possible ways to arrange themselves.

5! = 5 4

= 20 3

= 120 1 = 120 possible ways to arrange themselves if the doctor were to sit in either the left - most or right - most seat.

In either case there are 120 ways, so 120 + 120 = Total of 240 arrangements among the 6 people if the doctor sits in the aisle seat ( leftmost or rightmost seat )

c. With each husband on the left, there are 3 people left, all women, that we have to consider here.

3!  = 3 2 6 ways to arrange 3 couples in a row, the husband always to the left

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!

it would to have a graph or something ! but i had this question so its false

step-by-step explanation: