How many numbers divisible by 5 lying between 3000 and 4000 can be formed from the digits 3, 4, 5, 6, 7 and 8 no digit being repeated in any number?

Hint: First, we have to identify whether this problem is related to permutation or combination. As per the question, we have to make possible integers without any repetition of same number of this will be solved by using permutation given as ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ . So, order of the number will matter here.

Complete step by step answer:
In the question, we are asked to find those numbers which are lying between 3000 and 4000 which are divisible by 5. Also, those numbers should have digits like 3, 4, 5, 6, 7, 8 without being repeated.
So, in this order of numbers will matter. Thus, permutation will be used by using formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ .
Now, we can easily understand that numbers between 3000 and 4000 will have ${{1}^{st}}$ digit fix i.e. 3. So, now we have remaining 3 places to make 4-digit numbers divisible by 5. Also, we know that any number to be divisible by 5 should have either of the digit i.e. 0 or 5 at unit place. So, here we don’t have to use 0 so, last digit will be 5. Thus, now ${{1}^{st}}$ and last digit are fix i.e. 3 and 5 respectively.
Now, the middle two places are left to be filled. Thus, we have the remaining digit 4,6,7,8 to be used. So, total 4 digit and 2 places. Therefore, using permutation formula as
${}^{4}{{P}_{2}}=\dfrac{4!}{\left( 4-2 \right)!}$ where n is total digit to be used and r is the number of places to be filled.
$=\dfrac{4!}{2!}=\dfrac{4\times 3\times 2\times 1}{2\times 1}=12$
Thus, there are 12 possible ways to fill the middle two places.

So, 12 numbers are there between 3000 and 4000 divisible by 5 without repeating the digits.

Note: Another approach can be manually finding all the numbers which are divisible by 5 and then separating them into the digit 3,4,5,6,7,8 that can be very tedious and time consuming. Also, don’t get confused in using permutation formula as it is similar to combination i.e. given as ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$ .