JEE Main 2026 Apr 5 Shift 2, Mathematics Q9: Combinations

Treating the balls as distinct, partition 8 into (B, Y, R) with each colour at least 2. The valid splits are (2,2,4), (2,3,3), (2,4,2), (3,2,3), (3,3,2), (4,2,2).

Counting using $\left(\genfrac{}{}{0ex}{}{5}{b}\right)\left(\genfrac{}{}{0ex}{}{6}{y}\right)\left(\genfrac{}{}{0ex}{}{4}{r}\right)$ for each split:

$\left(\genfrac{}{}{0ex}{}{5}{2}\right)\left(\genfrac{}{}{0ex}{}{6}{2}\right)\left(\genfrac{}{}{0ex}{}{4}{4}\right)+\left(\genfrac{}{}{0ex}{}{5}{2}\right)\left(\genfrac{}{}{0ex}{}{6}{3}\right)\left(\genfrac{}{}{0ex}{}{4}{3}\right)+\left(\genfrac{}{}{0ex}{}{5}{2}\right)\left(\genfrac{}{}{0ex}{}{6}{4}\right)\left(\genfrac{}{}{0ex}{}{4}{2}\right)+\left(\genfrac{}{}{0ex}{}{5}{3}\right)\left(\genfrac{}{}{0ex}{}{6}{2}\right)\left(\genfrac{}{}{0ex}{}{4}{3}\right)+\left(\genfrac{}{}{0ex}{}{5}{3}\right)\left(\genfrac{}{}{0ex}{}{6}{3}\right)\left(\genfrac{}{}{0ex}{}{4}{2}\right)+\left(\genfrac{}{}{0ex}{}{5}{4}\right)\left(\genfrac{}{}{0ex}{}{6}{2}\right)\left(\genfrac{}{}{0ex}{}{4}{2}\right)$

$=150+800+900+600+1200+450=4100$.