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The number of ways of arranging $n$ 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 $200\%$, The value of $n$ is:









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Option(D) is correct

If $\textbf{n is even}$, then the number of boys should be equal to number of girls, let each be $a$.

⇒ $n =2a$

Then the number of arrangements,

$= 2×a!×a!$

If one more students is added, then number of arrangements,

$= a!× (a+1)!$

But this is $200\%$ more than the earlier

$⇒ 3(2×a!×a!) = a!×(a+1)!$

$⇒ a+1 = 6$ and $a =5$

$⇒ \textbf{n=10}$

But if $\textbf{n is odd}$, then number of arrangements:

$= a!(a+1)!$

Where, $n = 2a+1$

When one student is included, number of arrangements:

$= 2(a+1)!(a+1)!$

⇒ By the given condition,

$2(a+1) = 3$, which is not possible.

Edit: Thank you Aamir and Rashi for providing valuable insights related to question in the comments.

(7) Comment(s)

Avneet Singh

the question is amazing but only thing i didn't get that how $a+1$=$6$?

Avneet Singh

and why multiplied it by $3$ since it has increased $200%$?


the arrangements have been increased by 200% then why we have multiply by 3 to equalize it to the previous arrangements plz omit my this confusion thx response will be appreciated


lets assume that the condition before the addition is 100%.. then after addition it increases by 200%, in result we have we can multiply it by 3 as we know that the result is 300%..


I don't understand from the form when it included 1 student. Can you explain it for me?

Aamir Ansari

I don't think this solution is correct.

When you assumed $n$ is even and then mentioned 'if one more student is added': the arrangement should be $2(a)!(a+1)!$. because the added student can be either a boy or a girl. so 2 cases.

Also in later cases you didn't consider the above mentioned possibility.

Please notify if I'm wrong.


Aamir bhai I guess the solution is right. When assumed value of $n$ is even and then mentioned 'if one more student is added': the arrangements is CORRECTLY mentioned as, $(a)!(a+1)!$.

Arrangement discussed here are AFTER we have made the choice and not including the choice.

Factor of 2 is added because $a$ students can be switched with the other $a$ students, but once students are $a+1$ (be it boys or girls), there is only 1 way of arranging them (starting with $a+1$ number of students.).

In case I am not clear do let me know.