Question

The number of five digit numbers formed using the digits 0, 2, 2, 4, 4, 5 which are greater than 40000 is: -
(a) 84
(b) 90
(c) 72
(d) 60

Answer 1

Consider two cases. In the first case fix the digit 4 in the first place and find the number of ways to arrange 5 digits at the remaining four places. In the second case fix the digit 5 in the first place and find the number of ways to arrange 5 digits at the remaining four places. In case any digit is repeating, divide the number of arrangements by the factorial of the number of times that digit is repeated. Add the number of ways obtained in both cases to get the answer.

Complete step-by-step solution
Here, we have been provided with the digits 0, 2, 2, 4, 4, 5 and we have to find the total number of digit numbers that can be obtained using these given digits with the condition that the five-digit number should be greater than 40000. This condition will be possible only if the digit at the first place of the number will be 4 or 5. So, let us consider the following two cases: -
1. The digit in the first place is 4.

4| | | |
---|---|---|---|---

Here, the first place is fixed with digit 4, so now the remaining four places can be filled with the remaining five digits given as - 0, 2, 2, 4, and 5. So, we have,
Total number of ways to fill four boxes with five digits = $5\times 4\times 3\times 2$
Now, we can see that the digit 2 is repeating twice, so we have,
Total number of effective ways to fill four boxes = $\dfrac{5\times 4\times 3\times 2}{2!}=60$
2. The digit in the first place is 5.

5| | | |
---|---|---|---|---

Here, the first place is fixed with digit 5, so now the remaining four places can be filled with the remaining five digits given as - 0, 2, 2, 4, and 4. So, we have,
Total number of ways to fill four boxes with five digits = $5\times 4\times 3\times 2$
Now, we can see that the digits 2 and 4 are both repeating twice, so we have,
Total number of effective ways to fill four boxes = $\dfrac{5\times 4\times 3\times 2}{2!\times 2!}=30$
Therefore, the overall number of arrangements possible will be the sum of the effective number of arrangements possible in both cases. So, we have,
Number of five-digit numbers greater than 40000 = 60 + 30 = 90
Hence, option (b) is the correct answer.

Note: You may note that there are no other cases possible to form five-digit numbers greater than 40000. We cannot fill the first place with 0 or 2. Remember that if any digits are repeating ‘n’ times then we have to divide the total arrangements by n! to get the effective number of arrangements. Do not take the product of the number of ways obtained in both the cases.