Question

How many 6-digit whole numbers have the digits 2, 0, 2, and 2 as a 4-digit consecutively ordered block with no other digits between them?

Answer 1

There are 1680 digits 2, 0, 2, and 2 as a 4-digit consecutively ordered block with no other digits between them

How to determine the selection

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:

$Case\ 1 = 3 * 1 * 2 * 1 * 10 * 10$

$Case\ 1 = 600$

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:

$Case\ 2 = 9 * 3 * 1 * 2 * 1 * 10$

$Case\ 2 = 540$

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:

$Case\ 3 = 9 * 10 * 3 * 1 * 2 * 1$

$Case\ 3 = 540$

The total number of whole numbers is:

$Total = 600 + 540 + 540$

$Total = 1680$

Hence, there are 1680 6-digit whole numbers that the digits

Read more about combination and permutation at:

brainly.com/question/2292449