Question

A five digited number is formed by the digits $1,2,3,4,5,6,7$ and $8$. The probability that the number has even digit at both ends is

• A. $\frac{3}{14}$
• B. $\frac{3}{7}$
• C. $\frac{4}{7}$
• D. $\frac{5}{7}$

Answer 1

The correct option is A $\frac{3}{14}$

$E---E$
Selection of 2 even digits out of 4 even digits ${=}^4{C}_2$ and it's arrangement in 2! ways
Selection of 3 digits from remaining 6 digits ${=}^6{C}_3$ and it's arrangement in 3! ways
Total cases of selecting 5 digits out of 8 digits ${=}^8{C}_5$ and it's arrangement in 5! ways
Probability that the number has even digit at both ends is $=\frac{{}^4{C}_2\times 2!{\times }^6{C}_3\times 3!}{{}^8{C}_5\times 5!}=\frac{3}{14}$