Question

In a jeep there are 3 seats in front and 3 in the back.Number of different ways in which

Answer 1

In which.........what

Answer 2

The correct question: In a jeep there are 3 seat in front and three in the back, number of different ways in which six persons of different height can be seated so that every one in front is shorter than the person directly behind him.

a). 90

b). 120

c). 75

d). 60

The correct answer is option (a). 90.

Given:

Number of seats in front side = 3.

Number of seats in back side = 3.

Number of persons = 6.

To Find:

With the given data, we have to find the number of different ways in which six persons of different height can be seated so that every one in front is shorter than the person directly behind him.

Solution:

The 6 persons can be divided into 3 equal groups as follows:

$P_{1}\\P_{4}$     $P_{2}\\P_{5}$     $P_{3}\\P_{6}$

∴, The number of ways in which the 6 persons can be divided into 3 equal groups = $\frac{6!}{2!.2!.2!.3!} = \frac{6!}{(2!)^{3} .3!}.$

Let they be seated as where $P_{1}, P_{2}, P_{3}$  sitting on front seats are shorter than $P_{4}, P_{5}, P_{6}$ sitting on the back seats, respectively.

Then, $(P_{1}, P_{4}), (P_{2}, P_{5}), (P_{3}, P_{6})$ can be arranged in $3!$ ways.

∴, The possible cases are = $\frac{6!.3!}{(2!)^{3} . 3!}$ = $90.$

Hence, the number of different ways in which six persons of different height can be seated so that every one in front is shorter than the person directly behind him is 90.

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