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

In this problem we are asked to find in how many ways can six boys and five girls stand in a row if all the girls are to stand together with the condition that the boys cannot all stand together. We are going to solve this problem in two cases. Here we use simple multiplication and permutation conditions to solve this problem.

Complete step-by-step solution:
Total number of girls $= 5$
Number of ways $5$ girls stand together $= 5! = 120$
Total number of boys $= 6$
Number of ways $6$ boys stand together $= 6! = 720$
Let us consider the five girls as one unit. We also have six 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 $= 7!5! - - - - - \left( 1 \right)$
However, in $\left( 1 \right)$ above we have counted these two cases:
Case (i): All the boys stand together followed by all the girls.
Case (ii): All the girls stand together followed by all the boys.
But by the problem statement, these arrangements are disallowed.
The number of ways in which all boys stand together followed by all the girls $= 6!5! - - - - - \left( 2 \right)$
The number of ways in which all girls stand together followed by all the boys $= 5!6! - - - - - \left( 3 \right)$
So, from $\left( 2 \right){\text{ and }}\left( 3 \right)$ above, we have the number of disallowed arrangements $= 6!5! + 5!6! = 2\left( {6!5!} \right) - - - - - \left( 4 \right)$
Hence, the total number of ways of arrangements can be obtained from $\left( 1 \right)$ and $\left( 4 \right)$ as $\Rightarrow 7!5! - 2\left( {6!5!} \right)$
$\Rightarrow 7 \times 6!5! - 2\left( {6!5!} \right)$
Taking the common similar term to simplify,
$\Rightarrow 6!5!\left( {7 - 2} \right)$
Hence we get,
$\Rightarrow 6!5!5$
$\Rightarrow \left( {6 \times 5 \times 4 \times 3 \times 2 \times 1} \right)\left( {5 \times 4 \times 3 \times 2 \times 1} \right)5$
Multiply we get,
$\Rightarrow 432,000$

Therefore, Option B is the correct answer.

Note: There is an alternate method to solve this problem.
Total number of girls $= 5$
Number of ways $5$ girls stand together $= 5! = 120$
Total number of boys $= 6$
Number of ways $6$ boys stand together $= 6! = 720$
If all the girls stand together, there will be a total of $5$ places in between the $6$ boys to stand all the girls together as the girls cannot stand at any of the row.
$\therefore$Total number of ways if all the five girls stands together but all boys are not stand together $= 120 \times 720 \times 5 = 432,000$