Question

Determine the number of n-digit numbers with all digits odd, such that 1 and 3 each occur a nonzero, even number of times.

Answer 1

To determine the number of n-digit numbers with all odd digits, where 1 and 3 occur an even number of times and not zero times, we must use combinatorial methods involving permutations and combinations.

Step-by-step explanation:

The problem presented is a combinatorial mathematics problem that requires us to find the number of n-digit numbers where all digits are odd, with 1 and 3 each appearing an even number of times and not zero times. To solve this, we need to consider the number of ways to distribute the digits '1' and '3' into n positions, ensuring that both digits appear an even number of times. After planning the distribution of '1' and '3', we need to account for the placement of the remaining odd digits (5, 7, and 9) in the remaining positions.

It is essential to note that a systematic approach is required, where we could create patterns or formulas to calculate the required numbers efficiently. We’ll need to apply principles of permutation and combination, which could involve factorial calculations (e.g., 4! for four positions) and careful consideration of possible digit arrangements that meet the given criteria.

Answer 2

To determine the number of n-digit numbers with all digits odd, and 1 and 3 occurring a nonzero, even number of times, we can break it down into cases: when there are an even number of 1s and 3s combined, and when there are an odd number of 1s and 3s combined. The total number of n-digit numbers is 5^(n-2) + (5^(n-1) - 5^(n-3)).

Step-by-step explanation:

To determine the number of n-digit numbers with all digits odd, where 1 and 3 each occur a nonzero, even number of times, we can break it down into cases:

Case 1: There are an even number of 1s and 3s combined

In this case, we can choose any odd digit for each of the n digits, except for 1 and 3. There are 5 other odd digits to choose from (5, 7, 9, 5, 7, 9), resulting in 5n-2 possible numbers.

Case 2: There are an odd number of 1s and 3s combined

In this case, we need to have at least one digit as 1 or 3. The remaining (n-1) digits can be chosen as any odd digit (excluding 1 and 3), resulting in (5n-1 - 5n-3) possible numbers.

Therefore, the total number of n-digit numbers is 5n-2 + (5n-1 - 5n-3).