Answer to Question #146447 in Combinatorics | Number Theory for Josh

Question

In how many ways can 4 married couples attending a concert be seated in a row of eight seats if each couple must be seated together?
a.384
b.192
c.129
d.16

Answer 1

Let's observe one couple (two persons) as one object. Now we have 4 objects (couples). The number of permutations of 4 objects is

$$4!=24$$

Each couple consists of 2 persons. The number of permutations of 2 persons in each couple is

$$2!=2$$

First operation (the number of permutations of 4 couples) can be result in $n_1=24$ different ways.

Then for each of these ways a second operation (the number of permutations of persons in first couple) can be performed in $n_2=2$ different ways.

For each of these ways a third operation (the number of permutations of persons in second couple) can be performed in $n_3=2$ different ways.

For each of these ways a fourth operation (the number of permutations of persons in third couple) can be performed in $n_4=2$ different ways.

For each of these ways a fifth operation (the number of permutations of persons in fourth couple) can be performed in $n_5=2$ different ways.

The sequence of five operations can be performed in $24\cdot2\cdot2\cdot2\cdot2=384$ ways.

a.384