5 boys and 6 girls are arranged in all possib | vedclass

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

Question

(B) For $X$ (no two boys sit together): Arrange $6$ girls in $6!$ ways. There are $7$ gaps created: $\_ G \_ G \_ G \_ G \_ G \_ G \_$. We choose $5$

Vedclass · Accepted Answer

$21: 1: 1$

Vedclass · Answer

(B) For $X$ (no two boys sit together): Arrange $6$ girls in $6!$ ways. There are $7$ gaps created: $\_ G \_ G \_ G \_ G \_ G \_ G \_$. We choose $5$ gaps for $5$ boys in $^7C_5$ ways. So,$X = ^7C_5 	imes 5! 	imes 6! = 21 	imes 5! 	imes 6!$.For $Y$ (no two girls sit together): Arrange $5$ boys in $5!$ ways. There are $6$ gaps created: $\_ B \_ B \_ B \_ B \_ B \_$. We choose $6$ gaps for $6$ girls in $^6C_6$ ways. So,$Y = ^6C_6 	imes 5! 	imes 6! = 1 	imes 5! 	imes 6!$.For $Z$ (circular arrangement,no two boys sit together): Arrange $6$ girls in a circle in $(6-1)! = 5!$ ways. There are $6$ gaps between them. We choose $5$ gaps for $5$ boys in $^6C_5$ ways and arrange them in $5!$ ways. So,$Z = ^6C_5 	imes 5! 	imes 5! = 6 	imes 5! 	imes 5!$.Now,$X: Y: Z = (21 	imes 5! 	imes 6!) : (1 	imes 5! 	imes 6!) : (6 	imes 5! 	imes 5!)$.Dividing by $5! 	imes 5!$,we get $(21 	imes 6) : (1 	imes 6) : 6 = 126 : 6 : 6 = 21 : 1 : 1$.

Similar Questions

Find the remainder when $(1!)^2 + (2!)^2 + (3!)^2 + \dots + (100!)^2$ is divided by $10^2$.

The letters of the word $ASSASSIN$ are written down at random in a row. The probability that no two $S$ occur together is

The digit in the unit place of the number $(183!) + (3^{183})$ 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:

Let $m = (9n^2 + 54n + 80)(9n^2 + 45n + 54)(9n^2 + 36n + 35)$. The greatest positive integer which divides $m$ for all positive integers $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module