How many 8-digit numbers can be formed such t | vedclass

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(B) To form an $8$-digit number with distinct digits from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$,the first digit cannot be $0$.There are $9$ choic

Vedclass · Accepted Answer

$(9 	imes 9!) / 2$

Vedclass · Answer

(B) To form an $8$-digit number with distinct digits from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$,the first digit cannot be $0$.There are $9$ choices for the first digit (any digit from $1$ to $9$).For the remaining $7$ positions,we need to choose and arrange $7$ digits from the remaining $9$ available digits (including $0$).The number of ways to fill the remaining $7$ positions is $^9P_7$.Thus,the total number of $8$-digit numbers is $9 	imes ^9P_7$.Calculating this: $9 	imes \frac{9!}{(9-7)!} = 9 	imes \frac{9!}{2!} = \frac{9 	imes 9!}{2}$.

How many $8$-digit numbers can be formed such that all digits are distinct?

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