Question

How many 4-digit numbers are there that the leftmost digit is odd, the second digit is even, and all 4 digits are different?

a) 2,160
b) 2,880
c) 4,320
d) 5,040

Answer 1

The number of 4-digit numbers where the leftmost digit is odd, second is even, and all digits are different is calculated as 5 (first digit) × 5 (second digit) × 8 (third digit) × 7 (fourth digit) = 1,400, which does not match any of the provided answer options.

Step-by-step explanation:

To calculate the number of 4-digit numbers where the leftmost digit is odd, the second digit is even, and all 4 digits are different, we follow these steps:

• The first digit can be an odd number: 1, 3, 5, 7, or 9, so there are 5 possibilities.
• The second digit can be any even number between 0 and 8, giving us 5 possibilities again (0, 2, 4, 6, 8).
• For the third digit, we now have 8 choices because it can be any digit except the two we've already used.
• For the fourth digit, we have 7 remaining choices, avoiding the three previously selected digits.

Multiplying these together, we get the total number of combinations: 5 (odd choices) × 5 (even choices) × 8 × 7 = 1,400. Since none of the options a) 2,160 b) 2,880 c) 4,320 d) 5,040 matches this calculation, it appears there may be a mistake in either the question or the options provided.