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 "$INTERMEDIATE$" consists of $12$ letters: $I, N, T, E, R, M, E, D, I, A, T, E$.Vowels are: $I, E, E, I, A, E$ (total $6$ vowels: $3$ $E$

Vedclass · Accepted Answer

$\frac{6!}{2!} 	imes \frac{7!}{2!3!}$

Vedclass · Answer

(A) The word "$INTERMEDIATE$" consists of $12$ letters: $I, N, T, E, R, M, E, D, I, A, T, E$.Vowels are: $I, E, E, I, A, E$ (total $6$ vowels: $3$ $E$'s,$2$ $I$'s,$1$ $A$).Consonants are: $N, T, R, M, D, T$ (total $6$ consonants: $2$ $T$'s,$1$ $N, 1$ $R, 1$ $M, 1$ $D$).First,arrange the $6$ consonants. The number of ways is $\frac{6!}{2!}$.These $6$ consonants create $7$ gaps (including ends) where the $6$ vowels can be placed so that no two vowels are together.The number of ways to arrange the $6$ vowels in these $7$ gaps is $^7P_6$,but since there are repetitions among vowels ($3$ $E$'s and $2$ $I$'s),we divide by $3!2!$.Thus,the number of ways is $\frac{^7P_6}{3!2!} = \frac{7!}{3!2!}$.Total arrangements = $\frac{6!}{2!} 	imes \frac{7!}{3!2!}$.

The number of different words that can be formed from the letters of the word "$INTERMEDIATE$" such that two vowels never come together,is

Similar Questions

The number of different $5$-digit numbers greater than $50000$ that can be formed using the digits $0, 1, 2, 3, 4, 5, 6, 7$,such that the sum of their first and last digits is not more than $8$,is:

If $n$ is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero,then $n$ is equal to:

The total number of numbers lying between $100$ and $1000$ that can be formed with the digits $1, 2, 3, 4, 5$,if the repetition of digits is not allowed and the numbers are divisible by either $3$ or $5$,is:

If the letters of the word $ATTEMPT$ are rearranged randomly,what is the probability that all $T$'s occur together?

The number of all $3$-digit numbers $abc$ (in base $10$) for which $(a \times b \times c) + (a \times b) + (b \times c) + (c \times a) + a + b + c = 29$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module