Question

Consider digits $1,2,3,4,5,6$ and $7.$ Using these digits, numbers of five digits are formed. Then probability of these such five digit numbers that have odd digits at their both ends is

• A. $\frac{1}{7}$
• B. $\frac{2}{7}$
• C. $\frac{3}{7}$
• D. None of the options

Answer 1

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

The total five digits are formed from seven numbers
$={}^7{P}_5$

The total favorable conditions
= Number of ways to place odd digit at both ends $\times$ Number of ways of writing remianing 3 digits
$={}^4{P}_2\times {}^5{P}_3$

Required probability
$=\frac{{}^4{P}_2\times {}^5{P}_3}{{}^7{P}_5}=\frac{2}{7}$