Question

The number of ways in which 6 boys and 3 girls can sit together so that not all the girls come together is $A.332640$
B.323640$C.233640$
D.332640$$$$

Answer 1

We first calculate the total number of ways we can arrange all the boys and girls(9 persons ). Then we find the number of ways where the girls always sit together by treating the 3 girls as a single unit and then arranging within themselves. The difference between the total number of ways to the number of ways where girls sit together is the required result. $$$$

Complete step-by-step solution:
The factorial of the natural number $n$ is the product of the first $n$ natural number. We denote the factorial of $n$ as $n!$ and it is given by $n!=1\times 2\times 3\times ...\times n$ We also know from permutation that the number of ways we can arrange $n$ distinct objects into a particular order is given by $n!$.
We know the rule of product from the fundamental of counting that if there are $m$ ways to do something and $n$ ways to other things then there are $m\times n$ ways to do both things. It is given in the question that there are 6 boys and 3 girls who have to sit together. There are a total of $6+3=9$ number of persons. The boys and the girls are distinct, not identical objects. So we can arrange 9 numbers of persons in $9!$ ways.
Now we shall find the number ways all the girls sit together in the arrangement and then we can make the difference from the total number of ways to get the result. We first let the 6 boys sit leaving a gap of 3 seats between any two consecutive boys. We can fill the gap with the unit of 3 girls. Now we can arrange the 6 boys and the unit of 3 girls in $\left( 6+1 \right)!=7!$ ways. We can arrange the seated girls in $3!$ ways. So total number of ways where girls seat together is $7!\times 3!$$$$$ **Therefore the total number of ways where not all the girls come together is $9!-7!\times 3!=332640$. So the correct choice is A.**

Note: We note that we use the product rule because each boy and each girl is different from others. We differentiate it from the sum rule used when we have to either of the things If we make $n$ persons seat on the round table the total arrangement will be $\left( n-1 \right)!$.