Permutation-Combination
The number of ways of arranging nn students in a row such that no two boys sit together and no two girls sit together is m(m>100)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:
A. 1212
B. 88
C. 99
D. 10
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1 solution
If n is evenn is even, then the number of boys should be equal to number of girls, let each be aa.
⇒ n=2an=2a
Then the number of arrangements,
=2×a!×a!=2×a!×a!
If one more students is added, then number of arrangements,
=a!×(a+1)!=a!×(a+1)!
But this is 200%200% more than the earlier
⇒3(2×a!×a!)=a!×(a+1)!⇒3(2×a!×a!)=a!×(a+1)!
⇒a+1=6⇒a+1=6 and a=5a=5
⇒n=10⇒n=10
But if n is oddn is odd, then number of arrangements:
=a!(a+1)!=a!(a+1)!
Where, n=2a+1n=2a+1
When one student is included, number of arrangements:
=2(a+1)!(a+1)!=2(a+1)!(a+1)!
⇒ By the given condition,
2(a+1)=32(a+1)=3, which is not possible.