Five-digit numbers are formed using the digit | vedclass

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(A) The set of available digits is ${1, 2, 3, 4, 5, 6, 8}$. There are $7$ digits in total.Total number of $5$-digit numbers that can be formed using t

Vedclass · Accepted Answer

$\frac{2}{7}$

Vedclass · Answer

(A) The set of available digits is ${1, 2, 3, 4, 5, 6, 8}$. There are $7$ digits in total.Total number of $5$-digit numbers that can be formed using these $7$ digits (assuming repetition is not allowed) is $^7P_5 = 7 	imes 6 	imes 5 	imes 4 	imes 3 = 2520$.Even digits available in the set are ${2, 4, 6, 8}$,which are $4$ in total.For a $5$-digit number to have even digits at both ends,we must choose $2$ even digits for the two ends from the $4$ available even digits. The number of ways to arrange these is $^4P_2 = 4 	imes 3 = 12$.After placing $2$ even digits at the ends,we are left with $5$ digits (from the original $7$ digits minus the $2$ used). We need to fill the remaining $3$ positions using these $5$ digits. The number of ways to do this is $^5P_3 = 5 	imes 4 	imes 3 = 60$.Total favorable outcomes = $12 	imes 60 = 720$.Probability = $\frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{720}{2520} = \frac{72}{252} = \frac{2}{7}$.

Five-digit numbers are formed using the digits $1, 2, 3, 4, 5, 6$,and $8$. What is the probability that they have even digits at both the ends?

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