6 boys and 5 girls sit in a line such that (I | vedclass

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

Question

(C) Case $(I)$: No two girls sit together.We arrange $6$ boys in $6!$ ways. This creates $7$ gaps (including ends) where $5$ girls can be placed.Numbe

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Case $(I)$: No two girls sit together.We arrange $6$ boys in $6!$ ways. This creates $7$ gaps (including ends) where $5$ girls can be placed.Number of ways $p = 6! 	imes {}^{7}P_{5} = 6! 	imes \frac{7!}{2!} = 6! 	imes \frac{7 	imes 6 	imes 5 	imes 4 	imes 3}{1} = 6! 	imes 2520$.Case $(II)$: All girls sit together.Treat the $5$ girls as a single unit. Now we have $6$ boys and $1$ unit of girls,total $7$ entities.These $7$ entities can be arranged in $7!$ ways. The $5$ girls within the unit can be arranged in $5!$ ways.Number of ways $q = 7! 	imes 5!$.Ratio $p/q = \frac{6! 	imes \frac{7!}{2!}}{7! 	imes 5!} = \frac{6!}{2! 	imes 5!} = \frac{720}{2 	imes 120} = \frac{720}{240} = 3$.

$6$ boys and $5$ girls sit in a line such that $(I)$ no two girls sit together $(II)$ all the girls sit together. If $p$ is the number of arrangements in case $(I)$ and $q$ is the number of arrangements in case $(II)$,then $p/q =$

Similar Questions

How many $3$-digit odd numbers can be formed using the digits $1, 2, 3, 4, 5, 6, 7, 8, 9$ if repetition of digits is not allowed?

What is the number of permutations of three letters $A, B, C$ taken all at a time?

There are $11$ distinct books on a shelf,consisting of $5$ Mathematics,$4$ Physics,and $2$ Chemistry books. In how many ways can they be arranged such that books of the same subject are always kept together?

$10$ men and $6$ women are to be seated in a row so that no two women sit together. The number of ways they can be seated is:

The number of $4$ digit even numbers that can be formed using $0, 1, 2, 3, 4, 5, 6$ without repetition is

Vedclass Test Series

Exam Paper Generator

Online Exam Module