Number of ways = Total $-$ when $B_1 \& G_1$ sit together

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

Now, when $B_1 \& G_1$ 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 $7!-6!2!=5\times 6!$