Let's break down the problem into parts for clarity:

Part (a): How many 7-digit numbers have odd numbers in the odd positions?

The number is a 7-digit number, so it has the following positions:

• 1st position (odd)
• 2nd position (even)
• 3rd position (odd)
• 4th position (even)
• 5th position (odd)
• 6th position (even)
• 7th position (odd)

The odd positions are the 1st, 3rd, 5th, and 7th. For each of these positions, the digit must be an odd number. The odd digits are 1, 3, 5, 7, and 9, so there are 5 choices for each odd position.

For the even positions (2nd, 4th, 6th), there are no restrictions mentioned, so any digit (0 through 9) can be used, giving us 10 choices for each even position.

Calculation for part (a):

• Odd positions (1st, 3rd, 5th, 7th): 5 choices each (1, 3, 5, 7, 9)
• Even positions (2nd, 4th, 6th): 10 choices each (0 through 9)

Thus, the total number of such 7-digit numbers is:

$5 \times 5 \times 5 \times 5 \times 10 \times 10 \times 10 = 5^4 \times 10^3 = 625 \times 1000 = 625,000$

So, the number of 7-digit numbers with odd numbers in the odd positions is 625,000.

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Part (b): How many of these have even numbers in the even positions?

For the even positions (2nd, 4th, 6th), the digits must now be even numbers. The even digits are 0, 2, 4, 6, and 8, so there are 5 choices for each even position.

Therefore, for the odd positions (1st, 3rd, 5th, 7th), the calculation remains the same as in part (a), with 5 choices for each.

Calculation for part (b):

• Odd positions (1st, 3rd, 5th, 7th): 5 choices each (1, 3, 5, 7, 9)
• Even positions (2nd, 4th, 6th): 5 choices each (0, 2, 4, 6, 8)

Thus, the total number of such 7-digit numbers is:

$5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 = 5^7 = 78125$

So, the number of 7-digit numbers with odd numbers in the odd positions and even numbers in the even positions is 78,125.

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Summary:

a. 625,000 7-digit numbers have odd numbers in the odd positions.
b. 78,125 7-digit numbers have odd numbers in the odd positions and even numbers in the even positions.

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Would you like further explanations on any part of this? Or do you have additional questions?

Here are a few related questions to explore:

1. How many 7-digit numbers have digits restricted to odd numbers only?
2. What is the total number of 7-digit numbers with no restrictions?
3. How can we calculate the number of 7-digit numbers with no odd digits in odd positions?
4. How would the problem change if the first digit was restricted to being a non-zero odd number?
5. How can we generalize this type of counting problem for an $n$-digit number?

Tip: When working with positional constraints like this, always break the problem down by considering the different categories of positions (odd/even) and the possible choices for each.