Question

 Find the number of ways in which 5 boys and 4 girls can be arranged on a circular table such that no girls sit together and 2 particular boys are always together?
A.276
B.288
C.296
D.306

Answer 1

Hint: First of all, consider two boys that are together occupying a single place, then find the number of arrangements using the concept of circular permutation. We next find the number of arrangements when there are 4 girls and 4 places left between boys and no two girls can sit together. Next, multiply both the number of ways possible to find the number of ways in which 5 boys and 4 girls can be arranged on a circular table such that no girls sit together and 2 particular boys are always together.

Complete step-by-step answer:
Let us first assume that two boys that always together occupy one space.
Then, boys will take to have four places.
Since the boys will be arranged on a circular table.
It is known that the number of possible arrangements on a circular table is $\left( {n - 1} \right)!$, where $n$ is the number of total objects because when objects are arranged in a circular table, there is no specific starting point.
Also, those two boys can interchange their position.
Then, the number of ways boys can be arranged on a circular table is $\left( {4 - 1} \right)!2! = 3!2!$
Now, we will find the number of ways girls can be arranged.
There are 4 girls and there are 4 places left such that 4 girls can be arranged on a circular table such that no girls sit together, so there are 4! ways in which girls can be arranged.
Now, we have to calculate the total number of ways in which 5 boys and 4 girls can be arranged on a circular table such that no girls sit together and 2 particular boys are always together by multiplying the number of ways in both the cases.
Hence, we have $3!2! \times 4!$
Use the formula $n! = n.\left( {n - 1} \right)\left( {n - 2} \right).....3.2.1$, to solve the above expression.
$
\Rightarrow 3!2! \times 4! = 3.2.1.2.1.4.3.2.1 \\
 \Rightarrow 3!2! \times 4! = 288 \\
$
Hence, there are 288 ways possible.
Thus, option B is correct.

Note: Students must be careful about the term circular table used in the question. when $n$ things are arranged on a circular table, then there are $\left( {n - 1} \right)!$ possible ways. For better understanding, if there are 2 people who need to be arranged on a circular table, there is only one possible way. In a similar sense, if boys are to be arranged in 4 places, the number of arrangements is 4!