Question

Find the number of ways in which $5$-boys and $5$-girls can seat in a row so that all girls sit together.

Answer 1

Hint: First we have to define what the terms we need to solve the problem are.
Here for this kind of question we will use the concept of combination and permutation as we studied in our school days. And the given known things are there are $5$-boys and $5$-girls can sit in a row.
So , there are totally \[10\] persons in there and so that all girls sit together.

Complete step by step answer:
We are going to use permutation and combination methods to approach the given questions to find the number of ways, since the number of permutations of r-objects can be found from among n-things is ${}^n{p_r}$where p refers to the permutation. Also, similarly for combination we have r-things and among
n-things are ${}^n{c_r}$ and as we know the given things are $5$ boys and $5$ girls and a total of ten.
We will now permute by assuming all the five boys and five girls as one (because we have to make sure that all the boys and girls are sited together)
So, the number of ways in which can be done is ${}^2{p_2} = 2!$(acting permutation from the given formula)
Now we have the number of ways in which all the combinations of permuting a total of five girls, is ${}^5{p_5} = 5! = 120$ ways.
Therefore, number of ways in which $5$-boys and $5$-girls can seat in a row so that all girls sit together is
$86400$or $10! - 6(5)!$ (which is 10 factorials is total possible minus the possibility of boys is six times to found five factorial)

So, the correct answer is “Option A”.

Note: We also need to know some formula like; ${}^n{p_r} = \dfrac{{n!}}{{(n - r)!}}$for permutation
And also, the formula for combination is ${}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}}$the difference is one r factorial will be multiplied in denominator on the combination.