QJEEM24Maeb6b: JEE Main 2024 Mathematics - Class 11 JEEM24Maeb6b Solution: The total number of words (with or without meaning) that can be …

We have III, TT, D, S, R, B, U, O, N

Number of words with selection (a, a, a, b)

$={ }^8 \mathrm{C}_1 \times \frac{4 !}{3 !}=32$

Number of words with selection (a, a, b, b)

$=\frac{4 !}{2 ! 2 !}=6$

Number of words with selection (a, a, b, c)

$={ }^2 \mathrm{C}_1 \times{ }^8 \mathrm{C}_2 \times \frac{4 !}{2 !}=672$

Number of words with selection (a, b, c, d)

$\begin{aligned}& ={ }^9 \mathrm{C}_4 \times 4 !=3024 \\\\& \therefore \text {Total }=3024+672+6+32 \\\\& =3734\end{aligned}$