permutation and combination — MATHS STD 11 Science — Question

$13 !=2 \times 3 \times 4 \times 5 \times 6 \times 7\times 8 \times 9 \times 10 \times 11 \times 12 \times 13$

$=2^{10} \times 3^5 \times 5^2 \times 7 \times 11 \times 13$

$24^k=\left(2^3 \times 3\right)^k$

When $13\,!$ is divide by $24^k$

$\therefore 2^{10} \times 3^5 \times 5^2 \times 7 \times 11 \times 13$

$\quad 2^{3 k} \cdot 3^k$

$\quad=2^{10-3 k} \cdot 3^{5-k} \cdot 5^2 \times 7 \times 11 \times 13$

$\therefore \quad 10-3 k=\text { integer }$

Then, maximum value of $k=3$