Question

A five digit number is formed by the digits $1,2,3,4,5,6$ & $8$. Probability that the number has even digits at both the ends is-

• A. $3/7$
• B. $4/7$
• C. $2/7$
• D. $4/5$

Answer 1

Answer: (B)

$4/7$

$1,2,3,4,5,6$ & $8$

So ${}^7{P}_5=\mathrm{7.6.5.4.3}=2520$

There are $4$ even digits

$(2,4,6,8)$

$\therefore 2$ even digits can be selected in ${}^4{C}_2=6$ ways

$\therefore 2$ even digits can be $2$ ends in $2\times 6=12$ ways

Remaining $3$ places from remaining $6$ digits

${}^6{P}_3=\mathrm{6.5.4}=120$ ways

$\therefore$ Probability $=\frac{12\times 120}{2520}=\frac{4}{7}$