Total number of 6-digit numbers in which only | vedclass

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Question

(A) To form a $6$-digit number using the digits ${1, 3, 5, 7, 9}$,one digit must be repeated exactly twice,and the other four digits must appear exact

Vedclass · Accepted Answer

$\frac{5}{2}(6!)$

Vedclass · Answer

(A) To form a $6$-digit number using the digits ${1, 3, 5, 7, 9}$,one digit must be repeated exactly twice,and the other four digits must appear exactly once.Step $1$: Select the digit to be repeated. There are $5$ choices $(^{5}C_{1})$.Step $2$: Arrange these $6$ digits (where one digit is repeated twice). The number of arrangements is given by $\frac{6!}{2!}$.Step $3$: The total number of such $6$-digit numbers is $^{5}C_{1} 	imes \frac{6!}{2!} = 5 	imes \frac{720}{2} = 5 	imes 360 = 1800$.Note: $\frac{6!}{2!} = 360$,so $5 	imes 360 = 1800$. Comparing with the options,$\frac{5}{2}(6!) = 5 	imes 360 = 1800$.

Total number of $6$-digit numbers in which only and all the five digits $1, 3, 5, 7,$ and $9$ appear,is

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