Question

If $n$ persons are seated in a row, the number of ways of selecting $3$ persons so that no two of the mare seated next to each other is:

• A. $(n-2)C_3$
• B. $\frac{(n-2)!}{3!(n-5)!}$
• C. $(n-3)C_2+(n-3)C_3$
• D. $(n-3)C_3$

Answer 1

The correct option is B $\frac{(n-2)!}{3!(n-5)!}$

No. of ways of selecting $3$ persons out of $n$ person seated in a row so that no two of them are together is

$(n-2{)}_{C_3}=\frac{(n-2)!}{3!(n-2-3)!}=\frac{(n-2)!}{3!(n-5)!}$