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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.

(8) Comment(s)


No one said anything abt the number of boys or girls

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.