The number of different words that can be for | vedclass

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

Question

(A) The word $APPLICATION$ consists of $11$ letters: $A, P, P, L, I, C, A, T, I, O, N$.The vowels are $A, I, A, I, O$ ($5$ vowels) and the consonants

Vedclass · Accepted Answer

$(45)7!$

Vedclass · Answer

(A) The word $APPLICATION$ consists of $11$ letters: $A, P, P, L, I, C, A, T, I, O, N$.The vowels are $A, I, A, I, O$ ($5$ vowels) and the consonants are $P, P, L, C, T, N$ ($6$ consonants).First,arrange the $6$ consonants: $P, P, L, C, T, N$. The number of ways to arrange these is $\frac{6!}{2!} = 360$.There are $7$ possible gaps created by these $6$ consonants (including the ends) where the $5$ vowels can be placed so that no two vowels are together.The number of ways to choose $5$ gaps out of $7$ is $^7C_5 = \frac{7 	imes 6}{2} = 21$.The $5$ vowels $(A, A, I, I, O)$ can be arranged in these $5$ chosen gaps in $\frac{5!}{2!2!} = \frac{120}{4} = 30$ ways.Total number of words $= \frac{6!}{2!} 	imes ^7C_5 	imes \frac{5!}{2!2!} = 360 	imes 21 	imes 30 = 226800$.Calculating $(45)7! = 45 	imes 5040 = 226800$.Thus,the correct option is $(45)7!$.

The number of different words that can be formed using all the letters of the word $APPLICATION$ such that no two vowels come together is -

Similar Questions

The number of distinct primes dividing $12! + 13! + 14!$ is

The tens digit of the sum $1! + 4! + 7! + 10! + 12! + 13! + 16! + 17!$ is divisible by which of the following?

Words of length $10$ are formed by using the letters $A, B, C, D, E, F, G, H, I, J$. Let $x$ be the number of such words where no letter is repeated and $y$ be the number of such words where exactly two letters are repeated twice and no other letter is repeated,then the value of $\frac{y}{x}$ is

The number of $5$-digit odd numbers greater than $40,000$ that can be formed by using the digits $3, 4, 5, 6, 7, 0$ such that at least one of its digits is repeated is:

There are three sections in a question paper,each containing $4$ questions. If a candidate has to answer only $5$ questions from this paper without leaving any section,then the number of ways the candidate can make the choice of questions is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module