The number of 4 letter words (with or without | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The word $'EXAMINATION'$ contains $11$ letters: $A, A, I, I, N, N, E, X, M, T, O$. There are $8$ distinct letters: ${A, I, N, E, X, M, T, O}$. The

Vedclass · Accepted Answer

$2454$

Vedclass · Answer

(D) The word $'EXAMINATION'$ contains $11$ letters: $A, A, I, I, N, N, E, X, M, T, O$. There are $8$ distinct letters: ${A, I, N, E, X, M, T, O}$. The frequency of letters is: $A: 2, I: 2, N: 2, E: 1, X: 1, M: 1, T: 1, O: 1$. We need to form $4$ letter words. The cases are:$1$. Two letters of one kind and two letters of another kind:Selection: $^3C_2 = 3$ ways. Arrangement: $3 	imes \frac{4!}{2!2!} = 3 	imes 6 = 18$ ways.$2$. Two letters of one kind and two different letters:Selection: $^3C_1$ (for the pair) $	imes ^7C_2$ (for the two different letters) $= 3 	imes 21 = 63$ ways. Arrangement: $63 	imes \frac{4!}{2!} = 63 	imes 12 = 756$ ways.$3$. All four letters are different:Selection: $^8C_4 = 70$ ways. Arrangement: $70 	imes 4! = 70 	imes 24 = 1680$ ways.Total number of words $= 18 + 756 + 1680 = 2454$.

The number of $4$ letter words (with or without meaning) that can be formed from the eleven letters of the word $'EXAMINATION'$ is

Similar Questions

In a group of $6$ boys and $4$ girls,four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

$^{47}C_4 + \sum_{r=1}^{5} {}^{52-r}C_3 = $

Choose the correct number of ways in which $15$ different books can be divided into five heaps of equal number of books.

In how many ways can $15$ different books be divided equally among $5$ sets or groups?

Vedclass Test Series

Exam Paper Generator

Online Exam Module