The number of arrangements of the letters of | vedclass

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

Question

(C) The word $PALANHAR$ has $8$ letters: $P, A, L, A, N, H, A, R$. The vowels are $A, A, A$ and consonants are $P, L, N, H, R$. There are $5$ consonan

Vedclass · Accepted Answer

$2880$

Vedclass · Answer

(C) The word $PALANHAR$ has $8$ letters: $P, A, L, A, N, H, A, R$. The vowels are $A, A, A$ and consonants are $P, L, N, H, R$. There are $5$ consonants and $3$ vowels. Total positions are $8$ $(1, 2, 3, 4, 5, 6, 7, 8)$. Odd positions are $1, 3, 5, 7$ ($4$ positions) and even positions are $2, 4, 6, 8$ ($4$ positions). We need exactly two vowels at odd places and one vowel at an even place,such that no two vowels are together. First,arrange the $5$ consonants: $P, L, N, H, R$. They create $6$ gaps: $\_ C \_ C \_ C \_ C \_ C \_$. Number of ways to arrange consonants is $5! = 120$. We have $4$ odd positions and $4$ even positions. To have no two vowels together,we must place vowels in the gaps between consonants. After placing $5$ consonants,there are $6$ gaps. $3$ gaps are at odd positions and $3$ gaps are at even positions. We need $2$ vowels in odd positions and $1$ vowel in an even position. Number of ways to choose $2$ odd gaps from $4$ is $\binom{4}{2} = 6$. Number of ways to choose $1$ even gap from $4$ is $\binom{4}{1} = 4$. Total ways = $5! 	imes \binom{4}{2} 	imes \binom{4}{1} = 120 	imes 6 	imes 4 = 2880$.

The number of arrangements of the letters of the word $PALANHAR$ in which no two vowels are together and exactly two vowels are at odd places,is

Similar Questions

$A$ polygon has $54$ diagonals. Then,the number of its sides is

There are $12$ points in a plane,no three of which are in the same straight line,except $5$ points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these $12$ points is

How many triangles can be formed using $5$ points on a line and $3$ points on a parallel line?

In a regular $15$-sided polygon with all its diagonals drawn,a diagonal is chosen at random. The probability that it is neither a shortest diagonal nor a longest diagonal is

Let $t_n$ denote the number of integral-sided triangles with distinct sides chosen from $\{1, 2, 3, \ldots, n\}$. Then,$t_{20} - t_{19}$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module