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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) To form a $6$-digit number using all five 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 from the $5$ available digits. This can be done in $^5C_1 = 5$ ways.Step $2$: Arrange these $6$ digits (where one digit is repeated twice) in a sequence. The number of permutations is given by $\frac{6!}{2!}$.Step $3$: The total number of such $6$-digit numbers is $^5C_1 	imes \frac{6!}{2!} = 5 	imes \frac{720}{2} = 5 	imes 360 = 1800$.Note: $\frac{5}{2}(6!) = 2.5 	imes 720 = 1800$. Thus,the correct option is $A$.

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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