Question

The number of 4−letter words that can be formed out of the letters of the word "FIITJEE" is
A. 5040
B. 1260
C. 270
D. 180

Answer 1

In this question we use the theory of permutation and combination. So, before solving this question you need to first recall the basics of this chapter. For example, if we need to select two letters out of four. In this case, this can be done in ${}^{\text{4}}{{\text{C}}_2}$ =6 ways.
Formula use- ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$

Complete step-by-step answer:
Letters of word ‘FIITJEE’ consist of F, 2I’s, T, J and 2E’s.
We have to consider all possible combinations of 4 letters from the given word ‘FIITJEE’.
Now consider,
Case-1
1 I and 1 E
2 must be selected from T , J , E
No. of words = 4! $\times$3
= 72
Case-2
1 I and no E
All F, T, J must be in word
No. of words = 4!
=24
Case-3
1 E and no I
All F, T, J must be in word
No. of words = 4! = 24
Case-4
2 E’s and no I
2 letters selected from F, T, J must be in word
No. of words = $\dfrac{{4!}}{{2!}}$ $\times$ ${}^3{{\text{C}}_2}$
= 36
Case-5
2 I’s and no E
2 letters selected from F, T, J must be in word
No. of words = $\dfrac{{4!}}{{2!}}$ $\times$ ${}^3{{\text{C}}_2}$
= 36
Case-6
2 E’s and 1 I
No. of words = $\dfrac{{4!}}{{2!}}$ $\times$ ${}^3{{\text{C}}_2}$
= 36
Case-7
2 I’s and 1 E
No. of words = $\dfrac{{4!}}{{2!}}$ $\times$ ${}^3{{\text{C}}_2}$
= 36
Case-8
2 I’s and 2 E’s
No. of words = $\dfrac{{4!}}{{{{\left( {2!} \right)}^2}}}$
= 6
Thus, total no. of words = 72+24+24+36+36+36+36+6
= 270
Therefore, option (C) is the correct answer.

Note: We need to remember this formula for selecting r things out of n.
${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$
where n! means the factorial of n.
for example, $3!{\text{ = 3}} \times {\text{2}} \times 1$