A 5-digit number is formed using the digits 1 | vedclass

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(A) The set of available digits is $S = \{1, 2, 3, 4, 5, 6, 8\}$,which contains $7$ digits in total.Among these,the even digits are $\{2, 4, 6, 8\}$,s

Vedclass · Accepted Answer

$2/7$

Vedclass · Answer

(A) The set of available digits is $S = \{1, 2, 3, 4, 5, 6, 8\}$,which contains $7$ digits in total.Among these,the even digits are $\{2, 4, 6, 8\}$,so there are $4$ even digits and $3$ odd digits.The total number of $5$-digit numbers that can be formed using these $7$ digits is $P(7, 5) = \frac{7!}{2!} = 7 	imes 6 	imes 5 	imes 4 	imes 3 = 2520$.For the number to have even digits at both ends,we must choose $2$ even digits out of $4$ for the two ends and arrange them in $P(4, 2) = 4 	imes 3 = 12$ ways.The remaining $3$ positions can be filled by the remaining $5$ digits in $P(5, 3) = 5 	imes 4 	imes 3 = 60$ ways.Thus,the number of favorable outcomes is $12 	imes 60 = 720$.The probability is $\frac{720}{2520} = \frac{72}{252} = \frac{2}{7}$.

$A$ $5$-digit number is formed using the digits $1, 2, 3, 4, 5, 6,$ and $8$. What is the probability that the number has even digits at both ends?

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