How many numbers can be formed using the digi | vedclass

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Question

(B) To form a number between $3000$ and $4000$,the thousands place must be occupied by the digit $3$.For the number to be divisible by $5$,the units p

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) To form a number between $3000$ and $4000$,the thousands place must be occupied by the digit $3$.For the number to be divisible by $5$,the units place must be occupied by $0$ or $5$. Since $0$ is not in the given set of digits,the units place must be $5$.We have $6$ digits in total: ${3, 4, 5, 6, 7, 8}$.Fixing $3$ at the thousands place and $5$ at the units place,we have used $2$ digits.There are $4$ remaining digits ${4, 6, 7, 8}$ to fill the remaining $2$ places (hundreds and tens).The number of ways to arrange $4$ remaining digits in $2$ places is given by the permutation formula $^nP_r = \frac{n!}{(n-r)!}$.Here,$n = 4$ and $r = 2$,so $^4P_2 = \frac{4!}{(4-2)!} = \frac{4 	imes 3 	imes 2 	imes 1}{2 	imes 1} = 12$.Therefore,there are $12$ such numbers.

How many numbers can be formed using the digits $3, 4, 5, 6, 7, 8$ that lie between $3000$ and $4000$ and are divisible by $5$,given that repetition of digits is not allowed?

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