QUESTION

# There are 4 boys and 4 girls. In how many ways they can sit in row, not all girls sit together.

Hint:In this question firstly arrange all 8 people (boys and girls) in such a way that all the girls are together. The total arrangement for 8 people subtracted with the case of all the girls together gives the ways to sit them such that no girl sits together.

Given data
4 boys and 4 girls.
Now we have to find the number of ways so that no girl sits together.
As there are a total 8, boys and girls.
So the total number of ways to arrange them = (8!).
Now let us consider 4 girls as one group.
So the number of ways to arrange 4 boys and 1 group of girls (having in total 4 girls) will be the total arrangement of 4 girls in one group, and then we will have 5! In total as there are 4 boys and 1 group of girls that is 5, hence
$\Rightarrow 4! \times 5!$
So the total number of ways that no girl is together is the difference of total number of ways to arrange 8, boys and girls and the total number of ways in which all girls are together.
$\Rightarrow 8! - \left( {4! \times 5!} \right)$
Now simplify this we have,
$\Rightarrow \left( {8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \right) - \left( {4 \times 3 \times 2 \times 1 \times 5 \times 4 \times 3 \times 2 \times 1} \right)$
$\Rightarrow 40320 - 2880 = 37440$
So this is the required total number of ways so that no girls are together.

Note – Since there were 8 people to be seated thus the total ways were 8! As they can have permutations amongst themselves. The concept of considering certain people into a single group helps solving problems of this certain type, as making them sit together by making them a part of a group helps finding the ways not to make them sit together.