Five-digit numbers are formed using the digit | vedclass

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(A) The set of available digits is $S = \{1, 2, 3, 4, 5, 6, 8\}$. The total number of digits is $n = 7$.The total number of $5$-digit numbers that can

Vedclass · Accepted Answer

$\frac{2}{7}$

Vedclass · Answer

(A) The set of available digits is $S = \{1, 2, 3, 4, 5, 6, 8\}$. The total number of digits is $n = 7$.The total number of $5$-digit numbers that can be formed using these $7$ digits is $^7P_5 = \frac{7!}{2!} = 7 	imes 6 	imes 5 	imes 4 	imes 3 = 2520$.For the $5$-digit number to have even digits at both ends,we identify the even digits in the set: $\{2, 4, 6, 8\}$. There are $4$ even digits.The number of ways to fill the first and last positions with even digits is $^4P_2 = 4 	imes 3 = 12$.The remaining $3$ positions can be filled by the remaining $5$ digits in $^5P_3 = 5 	imes 4 	imes 3 = 60$ ways.Thus,the number of favorable outcomes is $12 	imes 60 = 720$.The probability is $\frac{720}{2520} = \frac{72}{252} = \frac{2}{7}$.

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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