In how many ways can the letters of the word | vedclass

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

Question

(A) The word $PERMUTATIONS$ contains $12$ letters: $P, E, R, M, U, T, A, T, I, O, N, S$. The letter $T$ repeats $2$ times.There are $12$ positions. If

Vedclass · Accepted Answer

$25401600$

Vedclass · Answer

(A) The word $PERMUTATIONS$ contains $12$ letters: $P, E, R, M, U, T, A, T, I, O, N, S$. The letter $T$ repeats $2$ times.There are $12$ positions. If we place $P$ at position $i$,then $S$ must be at position $i+5$ to have $4$ letters between them.The possible pairs of positions $(i, i+5)$ are $(1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11), (7, 12)$. There are $7$ such pairs.Since $P$ and $S$ can be interchanged,there are $7 	imes 2 = 14$ ways to place $P$ and $S$.The remaining $10$ letters can be arranged in $\frac{10!}{2!}$ ways because $T$ repeats twice.Total arrangements $= 14 	imes \frac{10!}{2!} = 7 	imes 10! = 7 	imes 3628800 = 25401600$.

In how many ways can the letters of the word $PERMUTATIONS$ be arranged if there are always $4$ letters between $P$ and $S$?

Similar Questions

Total number of four-digit odd numbers that can be formed using $0, 1, 2, 3, 5, 7$ (repetition is allowed) is:

How many $4$-digit even numbers can be formed using the digits $0, 1, 2, 3, 4, 5, 6$ without repetition?

Ten persons,amongst whom are $A, B$ and $C$,are to speak at a function. The number of ways in which it can be done if $A$ wants to speak before $B$ and $B$ wants to speak before $C$ is:

The total number of words that can be formed using the letters $a, b, c, d, e, f$ taken $3$ at a time such that each word contains at least one vowel is:

How many $5$-digit telephone numbers can be constructed using the digits $0$ to $9$ if each number starts with $67$ and no digit appears more than once?

Vedclass Test Series

Exam Paper Generator

Online Exam Module