Question

The number of ways of arranging $n+1$ students in a row such that no two boys sit together and no two girls sit together is $m(m>100).$ If one more student is added, then number of ways of arranging as above, increases by $400\%$. Then

• A. $n=17$
• B. $\frac{m}{n+1}=8!\times 9!$
• C. $n=18$
• D. $\frac{m}{n}=8!\times 9!$

Answer 1

Answer: (A), (B)

$\frac{m}{n+1}=8!\times 9!$

If $n+1$ is even, then the number of boys should be equal to number of girls, let each be $a$.
$\Rightarrow n+1=2a$
Then the number of arrangements,$=2\times a!\times a!$
If one more students is added, then number of arrangements,
$=a!\times (a+1)!$
But this is $400\%$ more than the earlier
$\Rightarrow 5(2\times a!\times a!)=a!\times (a+1)!\Rightarrow a+1=10\text{and}a=9\Rightarrow n=17$
But if $n+1$ is odd, then number of arrangements:
$=a!(a+1)!$
Where, $n+1=2a+1$
When one student is included, number of arrangements:
$=2(a+1)!(a+1)!$
$\Rightarrow$ By the given condition,
$2(a+1)=5$, which is not possible.
$\therefore m=2\times 9!\times 9!$