a. How many 7-digit numbers have odd numbers in the odd positions (i.e., odd numbers in the first, third, fifth, and seventh positions)?

To solve this, we need to count the number of valid 7-digit numbers where the digits in the odd positions (1st, 3rd, 5th, and 7th) are all odd.

Step-by-Step Breakdown:

1. Odd Positions (1st, 3rd, 5th, and 7th):

   • The odd digits are $1, 3, 5, 7, 9$, so there are 5 choices for each of these positions.

2. Even Positions (2nd, 4th, and 6th):

   • These positions can be any digit from $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$, so there are 10 choices for each of these positions.

Thus, the total number of 7-digit numbers with odd numbers in the odd positions is:

$5 \times 10 \times 5 \times 10 \times 5 \times 10 \times 5 = 5^4 \times 10^3$

$= 625 \times 1000 = 625,000$

b. How many of these have even numbers in the even positions?

Now, we are looking for 7-digit numbers where:

• The digits in the odd positions (1st, 3rd, 5th, 7th) are odd, as in part (a).
• The digits in the even positions (2nd, 4th, 6th) are even.

Step-by-Step Breakdown:

1. Odd Positions (1st, 3rd, 5th, and 7th):

   • These still must be odd numbers (1, 3, 5, 7, 9), so there are 5 choices for each of these positions.

2. Even Positions (2nd, 4th, and 6th):

   • The even digits are $0, 2, 4, 6, 8$, so there are 5 choices for each of these positions.

Thus, the total number of 7-digit numbers where odd digits are in the odd positions and even digits are in the even positions is:

$5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 = 5^7$

$= 78125$

Final Answers:

• a. The total number of 7-digit numbers with odd digits in the odd positions is $\mathbf{625,000}$.
• b. The total number of 7-digit numbers with odd digits in the odd positions and even digits in the even positions is $\mathbf{78,125}$.

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Would you like a deeper explanation or clarification on any part of the process?

Here are 5 related questions:

1. What is the total number of 7-digit numbers with all odd digits?
2. How would the count change if the number had 8 digits instead of 7, with the same conditions?
3. What if we allowed for the first digit to be zero in the original problem?
4. How many 7-digit numbers have even numbers in both the odd and even positions?
5. How does the number of choices in each position affect the total count of possible numbers?

Tip: Always break down the problem by positions (odd/even) and calculate the number of choices for each, then multiply them together to get the total number of possibilities!