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$\left( a \right)5 \times 6!$

$\left( b \right)5 \times 7!$

$\left( c \right)6 \times 6!$

$\left( d \right)7!$

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Hint: The number of ways in which $n$ people can be seated in a round table is $\left( {n - 1} \right)!$

Number of ways of arranging $5$ boys and $3$ girl, i.e. $8$ people on a round table would be $7!$

We subtract the number of ways of arranging those people when ${B_1}$ and ${G_1}$ are always together to get the required answer.

When ${B_1}$ and ${G_1}$are together, we get $4boys + 2girls + 1\left( {{B_1} + {G_1}} \right)$ i.e. $7$ people and since ${B_1} + {G_1}$ can be permuted in $2$ ways, these can be arranged in $6! \times 2$ ways.

Subtracting we have $7! - 6! \times 2 = 6!\left( {7 - 2} \right) = 5 \times 6!$ ways in total.

So, correct option is (A)

Note: Whenever we come across these types of problems first of all find number of ways of arranging $n$ people sitting in round table then we subtract the number of ways of arranging ${B_1}$ and ${G_1}$ are always together to get the required answer.

Number of ways of arranging $5$ boys and $3$ girl, i.e. $8$ people on a round table would be $7!$

We subtract the number of ways of arranging those people when ${B_1}$ and ${G_1}$ are always together to get the required answer.

When ${B_1}$ and ${G_1}$are together, we get $4boys + 2girls + 1\left( {{B_1} + {G_1}} \right)$ i.e. $7$ people and since ${B_1} + {G_1}$ can be permuted in $2$ ways, these can be arranged in $6! \times 2$ ways.

Subtracting we have $7! - 6! \times 2 = 6!\left( {7 - 2} \right) = 5 \times 6!$ ways in total.

So, correct option is (A)

Note: Whenever we come across these types of problems first of all find number of ways of arranging $n$ people sitting in round table then we subtract the number of ways of arranging ${B_1}$ and ${G_1}$ are always together to get the required answer.