JEE Main 2023 — Permutation Combination Question with Solution

According to the question,

$1, 1, 1, 2, 2, 3, 3, 4, 4$

there are four even digits as $2$ and $4$ are repeating two times each. 
So, the even digits will occupy the even places in $\frac{4!}{(2! 2!)} = 6$ ways

$3$ and $1$ are repeating two and three times respectively.
So, the remaining places will be occupied by odd digits in $\frac{5!}{(3! 2!)}=10$ ways

Therefore, total $9$ digits that can be formed $= 6\times 10 = 60$.
Hence, the answer is $60$.