Question

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

• A. $\frac{2}{7}$
• B. $\frac{3}{7}$
• C. $\frac{4}{7}$
• D. None

Answer 1

The correct option is A $\frac{2}{7}$

Total number of 5 digit numbers obtained by the digits 1, 2,3,4,5,6 and 8
${=}^7{P}_5=\mathrm{7.6.5.4.3}=2520$.
There are 4 even digits (2, 4, 6 and 8).
$\therefore$ 2 even digits can be selected in
${}^4{C}_2=6$ ways
$\therefore$ The two ends can be filled in
$2\times 6=12$ ways.
Remaining 3 places from remaining 5 digits can be filled in ${}^5{P}_3=60$ ways
Hence the required probability =$\frac{12\times 60}{2520}=\frac{2}{7}$
Short cut: