Menu Combinations and Gluten Items

Total combinations for Rohit $= \binom{16}{2} \times \binom{9}{3} = 120 \times 84 = 10080$. Let $V_{gf}$ be the number of gluten-free vegetarian items and $NV_{gf}$ be the number of gluten-free non-vegetarian items. Combinations for Bela $= \binom{V_{gf}}{2} \times \binom{NV_{gf}}{1} = \frac{V_{gf}(V_{gf}-1)}{2} \times NV_{gf}$. We are given $10080 = 12 \times \frac{V_{gf}(V_{gf}-1) NV_{gf}}{2} \implies V_{gf}(V_{gf}-1) NV_{gf} = 1680$. Since $V_{gf} \le 16$ and $NV_{gf} \le 9$, let's test possible integer values of $NV_{gf}$ (from 9 downwards): If $NV_{gf} = 8$: $V_{gf}(V_{gf}-1) = 1680 / 8 = 210 = 15 \times 14 \implies V_{gf} = 15$. Then total gluten-free items $= 15 + 8 = 23$. Since total menu items $= 16 + 9 = 25$, the number of items containing gluten $= 25 - 23 = 2$.