Conceptual math problem

Select where put $2$ l's $\rightarrow \binom{9}{2}$

Select where put $2$ i's $\rightarrow \binom{7}{2}$

Order other letters $5!$

$\boxed{\binom{9}{2}\binom{7}{2}\times 5!=90720}$

In the given word "BRILLIANT", there are 9 words.Since the letter I & L are repeated twice.

Therefore the number arrangements can be made out of the letter of the word "BRILLIANT" is $\frac {9!}{2! \times 2!}$ = $\boxed{90720}$

There are 9 letters with I and L appear two times. So the answer is $\frac{9!}{2!\times2!}=\boxed{90720}$

The answer is n!/p!xq! if any letter repeats p times and another letter repeats q times.

Here, n = 9 and p(L) = 2 and q(I) = 2.

So, answer = 9!/(2! x 2!) = 90720.

brilliant has 9 letters among which 2 letters are repeted 2times so \frac {9!} {2! \times 2!} = \boxed {90720}

Answer=[(9!)/{(2!)(2!)}]

9!/(2!*2!)