How many five-digit numbers are there such that the two leftmost digits are…

This is a restricted-arrangement counting problem, solved using the Fundamental Principle of Counting: when a number is built position by position with the choices at each position fixed, the total number of ways is the product of the choices available at every position; when one specific combined outcome across positions is expressly forbidden, that single case is removed by subtraction after the product is formed, rather than handled position by position.

1. The first digit (leftmost) of a five-digit number cannot be 0, and it must be even, so it can only be one of 2, 4, 6, 8 — 4 choices.

2. The second digit must also be even, and here 0 is allowed, so it can be one of 0, 2, 4, 6, 8 — 5 choices.

3. Multiplying these two positions freely gives 4 × 5 = 20 combinations for the leading pair, but exactly one of them has the digit 4 in both positions, which the question forbids as a repeated 4 — subtracting that single case leaves 20 − 1 = 19 valid combinations for the leading pair.

4. Each of the third, fourth and fifth digits must be odd, chosen independently from 1, 3, 5, 7, 9 — 5 choices at each of these three positions; since 4 is even, it can never occur here, so no further exclusion is needed.

5. By the counting principle, the total number of such five-digit numbers is the leading-pair count multiplied by the choices for the three trailing positions: 19 × 5 × 5 × 5 = 19 × 125 = 2375.

Cross-check: compute the count without any restriction on the leading pair first — 4 × 5 × 5 × 5 × 5 = 2500 — then subtract only the forbidden block where both leading digits are 4, i.e. 1 × 5 × 5 × 5 = 125 for the three free odd positions. 2500 − 125 = 2375, confirming the result by a second, independent route.

Hence, there are 2375 such five-digit numbers, matching option 2375.