Question

Number of 5-digit numbers which are divisible by 5 and each number containing the digit 5, digits being all different is equal to k(4!), the value of k is
(A) 84
(B) 168
(C) 188
(D) 208

Answer 1

Hint:
Permutations: Each of the arrangements which can be made by taking some or all number of things is called a permutation.
Here we use permutation without repetition to calculate the total numbers that can be formed.

Complete step by step solution:
We are given that each number contains 5 digits out of which one digit is compulsorily 5.
Now we require remaining 4 digits which can be selected from 9 digits, i.e. 0, 1, 2, 3, 4, 6, 7, 8, 9.
Since the numbers are divisible by 5 so unit’s digits can only be either 0 or 5.
I. Now, let us suppose that 0 is not one of the digits of the remaining 4 digits.
This implies that the unit’s digit must be 5 (otherwise the number will not be divisible by 5).
Now, since the unit’s digit is 5 and 0 is not included so number of choices to select the remaining 4 digits becomes $8\times 7 \times 6 \times 5$.
II. Now, let us suppose the case when 0 is included.
If 0 is included then unit’s digit can be either 0 or 5.

Case 1: Assuming 0 to be the unit’s digit.
In this case, 5 can be put in any one of the remaining 4 digits.
Then, in the remaining 3 digits, we have a choice of 8 numbers to be filled and those 3 places can be filled in $8\times 7 \times 6$ ways. Thus, total arrangements by considering the order, in this case, become $4 \times 8 \times 7 \times 6$.

Case 2: Assuming 5 to be the unit’s digit.
In this case, 0 cannot be put in the first place, as this number becomes a 4-digit number. So, for the first digit, we are left with 8 choices (1, 2, 3, 4, 6, 7, 8, 9).
Now, we can put 0 in any of the remaining 3 places. Suppose we fix 0 at second place, then, the remaining 2 places can be filled in $7\times 6$ ways.
Thus, the total number of arrangements by considering the order, in this case, becomes $8\times 3 \times 7 \times 6$.
So, total number of ways become: $8\times 7 \times 6 \times 5 +4 \times 8 \times 7 \times 6 +8 \times 3 \times 7 \times 6 =8 \times 7 \times 6 \times 12=4 \times 3 \times 2 \times 8 \times 7 \times 3=4! \times 168$
As total number of ways equals k(4!), thus,
k(4!)=168(4!)
On comparing, we get k = 168.

Hence, option (B) is correct.

Note:
In questions involving permutation without repetition, we have to reduce the number of choices available each time. If for the first time, we have let n choices then at the second time we are left with only (n-1) choices.