How many numbers can be made with the digits 3, 4, 5, 6, 7, 8 lying between 3000 and 4000 which are divisible by 5 when repetition of any digit is not allowed in any number?
A. 60
B. 12
C. 120
D. 24

Hint: Here in this question, we have to find the number of digits can be formed and that should be divisible by 5. To find the number of ways we use permutation and combination concept. Since this question involves arrangement of numbers, so we use the permutation concept.

Complete step by step solution:  An arrangement of items in a specific order is referred to as a permutation. Here, the components of sets are arranged in a linear or sequential order. While the order of the ingredients doesn't important in combination, it should be followed while performing permutations.
The 4 digits has to be formed from the given digits i.e., 3, 4, 5, 6, 7 and 8.
The number should be divisible by 5, so the unit places will be 5.
Since the number has to lie between 3000 and 4000, it has to start with 3.
Now the remaining numbers will occupy the remaining places. The remaining numbers are 4, 6, 7 and 8. So we have to select 2 numbers from the four numbers.
Therefore, the number of ways will be \[{}^{4}{{P}_{2}}\] ways = \[\dfrac{4!}{2!}\]ways
\[=12\] ways.
Therefore 12 numbers can be made with the digits 3, 4, 5, 6, 7 and 8 lying between 3000 and 4000 which are divisible by 5.
Hence option B is the correct one.

Note: The student should know the formula for the permutation and combination. The formula for the permutation is \[{}^{n}{{P}_{r}}=\dfrac{n!}{r!}\] and the formula for the combination is \[{}^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}\]. When solving permutation problems, we know about the factorial; it means that the function multiplies a number by every number below it.