The number of ways in which 5 boys and 3 girls can be seated on a...

Number of ways = Total $-$ when ${B}_{1} \text{ and } {G}_{1}$ sit together

Total ways in which $5\text{boys} + 3\text{girls} = 8$ people can be arranged is $7 !$

Now, when ${B}_{1} \text{ and } {G}_{1}$ sit together the number of ways (other $6$and ${B}_{1}{G}_{1}$ grouped together) is $6 ! \times 2 !$

Hence, the required numer of ways is $7 ! - 6 ! 2 ! = 5 \times 6 !$