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Question:

If $\frac{(2 n) !}{(3 !) \times(2 n-3) !}: \frac{n !}{(2 !) \times(n-2) !}=44: 3$, find the value of $n$

Solution:

Given Equation :

$\frac{(2 n) !}{(3 !) \times(2 n-3) !}: \frac{n !}{(2 !) \times(n-2) !}=44: 3$

To Find : Value of n

Formula: $n !=n \times(n-1) !$

By given equation

$\frac{(2 n) !}{(3 !) \times(2 n-3) !}: \frac{n !}{(2 !) \times(n-2) !}=44: 3$

$\therefore \frac{(2 n) !}{\frac{(3 !) \times(2 n-3) !}{n !}}=\frac{44}{3}$

$\therefore \frac{(2 n) !}{(3 !) \times(2 n-3) !} \times \frac{(2 !) \times(n-2) !}{n !}=\frac{44}{3}$

By using above formula

$\therefore \frac{(2 n) \times(2 n-1) \times(2 n-2) \times(2 n-3) !}{(3 \times 2 !) \times(2 n-3) !} \times \frac{(2 !) \times(n-2) !}{n \times(n-1) \times(n-2) !}$

$=\frac{44}{3}$

Cancelling terms $(n-2) !,(2 !),(2 n-3) ! \& n$, we get,

$\therefore \frac{2 \times(2 n-1) \times 2(n-1)}{3} \times \frac{1}{(n-1)}=\frac{44}{3}$

taking 2 common from the term $(2 n-2)$

$\therefore(2 n-1)=\frac{44 \times 3}{3 \times 2 \times 2}$

$\therefore(2 n-1)=11$

$\therefore n=6$

Conclusion : Value of n is 6.