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
594.6k+ views
Hint: 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.
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.
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.
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.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

In what year Guru Nanak Dev ji was born A15 April 1469 class 11 social science CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

