Question

Out of the first 30 natural numbers (i.e., 1 to 30), three are selected at random. The probability that no two are together

• A. $\frac{144}{145}$
• B. $\frac{28}{145}$
• C. $\frac{117}{145}$
• D. $\frac{2}{145}$

Answer 1

Answer: (C)

$\frac{117}{145}$

Select three numbers A, B and C. There are 27 numbers left. Insert a number between A and B and a number between B and C. This ensures that they are not together. There are 25 numbers left. These 25 numbers may be distributed in the 4 gaps (on selection of 3 numbers, there are 4 gaps created) in $\frac{{}^{(25+4-1)}{C}_{(4-1)}}{{}^{30}{C}_3}=\frac{{}^{28}{C}_3}{{}^{30}{C}_3}=\frac{28\ast 27\ast 26}{30\ast 29\ast 28}=\frac{117}{145}$