Words with or without meaning are to be forme | vedclass

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

Question

(D) The word $EXAMINATION$ contains $11$ letters: $A, A, E, I, I, M, M, N, N, O, T$.The total number of arrangements of these $11$ letters is given by

Vedclass · Accepted Answer

$\frac{1}{11}$

Vedclass · Answer

(D) The word $EXAMINATION$ contains $11$ letters: $A, A, E, I, I, M, M, N, N, O, T$.The total number of arrangements of these $11$ letters is given by $n(S) = \frac{11!}{2! 2! 2!}$,where $2!$ accounts for the repetitions of $A, I, M,$ and $N$.To find the number of arrangements where $M$ is fixed at the fourth position,we fix one $M$ at the fourth spot and arrange the remaining $10$ letters $(A, A, E, I, I, M, N, N, O, T)$.The number of such arrangements is $n(A) = \frac{10!}{2! 2! 2!}$.The probability is given by $P(A) = \frac{n(A)}{n(S)} = \frac{\frac{10!}{2! 2! 2!}}{\frac{11!}{2! 2! 2!}} = \frac{10!}{11!} = \frac{1}{11}$.

Words with or without meaning are to be formed using all the letters of the word $EXAMINATION$. The probability that the letter $M$ appears at the fourth position in any such word is:

Similar Questions

For which value of $n \in N$,does $n!$ have $13$ trailing zeros?

There are $4$ notes of Rs. $100$ and $5$ other notes of denominations Rs. $1$,Rs. $2$,Rs. $5$,Rs. $20$,and Rs. $50$. These $9$ notes are to be distributed among $3$ children such that each child receives at least one note of Rs. $100$. Find the total number of ways of distribution.

The value of $(2 \cdot {}^{1}P_{0} - 3 \cdot {}^{2}P_{1} + 4 \cdot {}^{3}P_{2} - \dots \text{ up to } 51^{\text{th}} \text{ term}) + (1! - 2! + 3! - \dots \text{ up to } 51^{\text{th}} \text{ term})$ is equal to

The number of $5$ digit telephone numbers having at least one of their digits repeated is

$^{80}C_{40}$ is not divisible by -

Vedclass Test Series

Exam Paper Generator

Online Exam Module