The number of ways, in which 5 girls and 7 boys can be seated at a round table so that no two girls sit together is

We need to seat 5 girls and 7 boys at a round table such that no two girls sit together.

Seating the 7 boys around a circle can be done in $$(7-1)! = 6! = 720$$ ways. This arrangement creates 7 gaps between consecutive boys.

We select 5 of these gaps for the girls in $$\binom{7}{5} = 21$$ ways. Then the 5 girls can be arranged among themselves in $$5! = 120$$ ways.

Multiplying these numbers gives the total arrangements as $$720 \times 21 \times 120 = 1814400$$.

This equals $$126 \times (5!)^2 = 126 \times 14400 = 1814400$$. Noting that $$\binom{7}{5} \times 6! = 21 \times 720 = 15120 = 126 \times 120 = 126 \times 5!$$, the answer can be succinctly written as $$126(5!)^2$$.