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Question

It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

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Solution

It is given that 5 men and 4 women are seated in a row. Since there are 5 men that are to be seated, thus the number of ways that the men are seated is the number of combination of 5 men taken all at a time.

The formula for permutation is,

P n r = n! ( nr )!

Substitute 5 for n and 5 for r in the above equation.

P 5 5 = 5! ( 55 )! = 5! 0! = 5! 1 =5!

The formula to calculate n! is defined as,

n!=1×2×3××( n1 )×n

Thus the factorial of 5 is,

5!=5×4×3×2×1 =120

Thus, the number of ways that the men are seated in the row is 120.

Now, since the women are to be seated in such a way that they occupy even seats, thus they have to seat between two men such that they can occupy even places. The number of ways that the women are seated is the number of combination of 4 women taken all at a time which will be calculated as 4! .

The formula to calculate n! is defined as,

n!=1×2×3××( n1 )×n

Thus the factorial of 4 is,

4!=4×3×2×1 =24

Thus the number of ways that the women are seated in the row is 24.

By multiplication principle which states that if an event can occur in m different ways and follows another event that can occur in n different ways, the possible number of arrangements are 120×24=2880 .


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