Question

The number of ways of rearranging the letter of the word “INSTITUTE”.

Answer 1

This is a problem of permutations and combinations and this problem just requires the concept of permutations. Permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor.
Apart from this we should have an idea of factorial.
We have the simple formula i.e.${n_{{P_r}}}$= $\dfrac{{n!}}{{\left( {n - r} \right)!}}$ is the numbers of ways of arranging r things from things.
This is also the definition of permutation that permutations are the different ways in which a collection of items can be arranged.

Complete step by step solution:
GIVEN:
The word i.e. “INSTITUTE” have 9 letters in which ‘I’ is repeated 2 times, ‘T’ is repeated 3 times means we have 2 I’s , 3 T’s , 1 N, 1 S, 1 U, 1 E i.e. total 9 letters.(2+3+1+1+1+1=9)
Therefore, no of ways of rearranging the letter of the word “INSTITUTE” = $\dfrac{{9!}}{{2!3!}}$
=$\dfrac{{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{(2 \times 1)(3 \times 2 \times 1)}}$
=30240
Here, we have developed this approach as in (factorial 9) ways we can rearrange the letters to form several words but some words will repeat as we have 2 I’s , 3 T’s so to avoid repetition we have divided (factorial 3) and (factorial 2) to (factorial 9) so that we finally get 30240 different words.

ANSWER - 30240 Ways

Note: In this type of question, we simply use the concept that if n things are to be arranged out of which ${r_1}$ things are of first type, ${r_2}$ things are of second type and ${r_3}$ things are of third type then:
$Total\,\,number\,\,of\,\,arrangements\, = \,\dfrac{{n!}}{{{r_1}!\,{r_2}\,!{r_3}\,!}}$