Question

In how many ways can six boys and five girls stand in a row if all the girls are to stand together but the boys cannot all stand together?
(a). 172,800
(b). 432,000
(c). 86,400
(d). None of these

Answer 1

Hint: The factorial is the product of all integers less than or equal to n but greater than or equal to 1. In this question, the five girls are considered as a one unit and the girls themselves can be arranged in 5! Ways.

Complete step-by-step answer:

Let us consider the 5 girls as one unit. We also have 6 boys. Hence, we now have 6 + 1 = 7 units that have to be arranged in a row. This can be done in 7! ways. But the girls themselves can be arranged in 5! ways.
Therefore, the total number of ways in which these 6 boys and 5 girls can be arranged in a row such that all the girls are together = \[7!\times 5!\] ……….. (1)
However, in the equation (1) above we have also counted these two cases:
All the boys stand together followed by all the girls
All the girls stand together followed by all the girls

By the problem statement, these arrangements are not allowed.
The number of ways in which all boys stand together followed by all the girls = \[6!\times 5!\] … (2)
The number of ways in which all girls stand together followed by all the boys = \[5!\times 6!\] … (3)
So, from the equation (2) and equation (3),
We have the number of not allowed arrangements = \[6!\times 5!\] + \[5!\times 6!\]= \[2\left( 6!\times 5! \right)\] …..(4)
The total number of required arrangements can be obtained from (1) and (4), we get
The required arrangements = $7!\times 5!-2\left( 6!\times 5! \right)$
The required arrangements = \[=6!\times 5!(7-2)\]
The required arrangements = \[=6!\times 5!\times 5\]
The required arrangements = \[=720\times 120\times 5=432000\]
Hence the correct option of the given question is option (b).

Note: Alternatively, the question is solved as follows. The boys can be arranging in 6! ways. The girls seating position is between the boys so boys cannot come together in 5 ways and the girls can arrange by these positions in 5! ways. Hence the total arrangement is $6!\times 5!\times 5=432000$ ways.