Simplify the expression. 5k(-6k^2 -2k+6)
1 answer:
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There are 1680 digits 2, 0, 2, and 2 as a 4-digit consecutively ordered block with no other digits between them
<h3>How to determine the selection</h3>To determine the number of whole numbers, the following must be true
Case 1: If the sequence starts from the first digit
- The first digit can be any of the three 2's (i.e. 3 digits)
- The second digit can only be 0
- The third digit can be any of the remaining 2's (i.e. 2 digits)
- The fourth digit can only be the last 2 (i.e. 1 digit)
- The fifth digit can be any of 0 - 9 (i.e. 10 digits)
- The sixth digit can be any of 0 - 9 (i.e. 10 digits)
So, we have:
Case 2: If the sequence starts from the second digit
- The first digit can be any of 1 - 9 (i.e. 9 digits)
- The second digit can be any of the three 2's (i.e. 3 digits)
- The third digit can only be 0
- The fourth digit can be any of the remaining 2's (i.e. 2 digits)
- The fifth digit can only be the last 2 (i.e. 1 digit)
- The last digit can be any of 0 - 9 (i.e. 10 digits)
So, we have:
Case 2: If the sequence starts from the third digit
- The first digit can be any of 1 - 9 (i.e. 9 digits)
- The second digit can be any of 0 - 9 (i.e. 10 digits)
- The third digit can be any of the three 2's (i.e. 3 digits)
- The fourth digit can only be 0
- The fifth digit can be any of the remaining 2's (i.e. 2 digits)
- The last digit can only be the last 2 (i.e. 1 digit)
So, we have:
The total number of whole numbers is:
Hence, there are 1680 6-digit whole numbers that the digits
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