JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Permutations and Combinations

The nine digits $1,1,2,2,2,2,3,4,4$ include three odd digits $1,1,3$, which must occupy the four even-numbered places. Choose which three of the four even places they take in $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ ways and arrange $1,1,3$ there in $\frac{3!}{2!}=3$ ways. The remaining six digits $2,2,2,2,4,4$ fill the other six places in $\frac{6!}{4!2!}=15$ ways. Multiplying, $4\times 3\times 15=180$ such numbers.