Find the number of arrangements of the letter | vedclass

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

Question

(B) The word $INDEPENDENCE$ has $12$ letters: $I(1), N(3), D(2), E(4), P(1), C(1)$.Total arrangements $= \frac{12!}{3! 	imes 2! 	imes 4!} = \frac{47

Vedclass · Accepted Answer

$1646400$

Vedclass · Answer

(B) The word $INDEPENDENCE$ has $12$ letters: $I(1), N(3), D(2), E(4), P(1), C(1)$.Total arrangements $= \frac{12!}{3! 	imes 2! 	imes 4!} = \frac{479001600}{6 	imes 2 	imes 24} = 1663200$.Treating all vowels $(I, E, E, E, E)$ as one unit,we have $8$ consonants $(N, N, N, D, D, P, C)$ and $1$ vowel unit. Total units $= 9$.Arrangements with vowels together $= \frac{9!}{3! 	imes 2!} 	imes \frac{5!}{4!} = \frac{362880}{6 	imes 2} 	imes 5 = 30240 	imes 5 = 151200$.Number of arrangements where vowels never occur together $= 1663200 - 151200 = 1512000$.

Find the number of arrangements of the letters of the word $INDEPENDENCE$. In how many of these arrangements do the vowels never occur together?

Similar Questions

While dialing a telephone number,an old man forgets the last two digits,remembering only that these are different. If he dials the last two digits at random,what is the probability that the number is dialed correctly?

Evaluate $\frac{n!}{(n-r)!}$,when $n=9, r=5$.

The number of different permutations that can be formed by taking $4$ letters at a time from the letters of the word '$REPETITION$' is

The number of ways in which $5$ boys and $3$ girls can be seated in a row so that each girl is between two boys.

How many numbers less than $10000$ can be formed using the eight digits $1, 2, 3, 4, 5, 6, 7, 0$? (Repetition of digits is allowed)

Vedclass Test Series

Exam Paper Generator

Online Exam Module