Find the number of ways, the digits 0,1,2 and 3 can be permuted to give rise to a number greater than 2000.
Please I need an explanatory workings!!!
You have to have the 2 or 3 as the first digit.
After that, the other 3 digits can be done in 3! = 6 ways.
So, the total is 2*3! = 12
2013
2031
2130
2310
2103
2301
3102
3120
3210
3201
3012
3021so therefore, 12ways.
A tree diagram can also be used to solve this problem.
To find the number of ways the digits 0, 1, 2, and 3 can be permuted to give a number greater than 2000, we can use a tree diagram.
First, let's consider the first digit. It must be either 2 or 3 to ensure the number is greater than 2000.
If the first digit is 2:
- The second digit can be any of the remaining 3 digits (0, 1, 3).
- The third digit can be any of the remaining 2 digits.
- The fourth digit can be the remaining digit.
This gives us a total of 3! = 6 permutations.
If the first digit is 3:
- The second digit can be any of the remaining 3 digits.
- The third digit can be any of the remaining 2 digits.
- The fourth digit can be the remaining digit.
Again, this gives us a total of 3! = 6 permutations.
Therefore, the total number of permutations that give rise to a number greater than 2000 is 6 + 6 = 12.
The answer for this question under permutations is 12
Abubakar
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To find the number of ways the digits 0, 1, 2, and 3 can be permuted to give rise to a number greater than 2000, we can use the method of counting.
1. Counting the number of ways the leftmost digit can be chosen:
- Since the number needs to be greater than 2000, the leftmost digit cannot be 0. Hence, we have 3 choices for the leftmost digit: 1, 2, or 3.
2. Counting the number of ways the second digit can be chosen:
- After choosing the leftmost digit, we have 3 remaining digits: 0, 2, and 3.
- If the leftmost digit is 1, then the second digit can be any of the 3 remaining digits (0, 2, or 3).
- If the leftmost digit is 2 or 3, then the second digit can be any of the 2 remaining digits (0 or 3).
- Therefore, the total number of choices for the second digit is:
- If the leftmost digit is 1: 3 choices
- If the leftmost digit is 2 or 3: 2 choices
3. Counting the number of ways the remaining digits can be chosen:
- After choosing the leftmost digit and the second digit, we have 2 remaining digits: 0 and the remaining digit that was not chosen for the leftmost digit and second digit.
- The remaining digits can be permuted in (2!) = 2 ways.
4. Combining all the choices to find the total number of ways:
- If the leftmost digit is 1:
- Number of choices for the leftmost digit: 1 choice (1)
- Number of choices for the second digit: 3 choices (0, 2, or 3)
- Number of choices for the remaining digits: 2! = 2 ways
- Total number of ways: 1 x 3 x 2 = 6 ways
- If the leftmost digit is 2 or 3:
- Number of choices for the leftmost digit: 2 choices (2 or 3)
- Number of choices for the second digit: 2 choices (0 or 3)
- Number of choices for the remaining digits: 2! = 2 ways
- Total number of ways for each case: 2 x 2 x 2 = 8 ways
Total number of ways: 6 (when the leftmost digit is 1) + 8 (when the leftmost digit is 2 or 3) = 14 ways
Therefore, there are 14 ways the digits 0, 1, 2, and 3 can be permuted to give rise to a number greater than 2000.