Multiple choice

How many natural numbers can be formed with the digits 4, 5, 8, 9, 3 and 2 (taken all together), so that all the even digits occupy even places?

  1. 6

  2. 12

  3. 18

  4. 36

  5. 54
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D Correct answer
Explanation

Digits: 4, 5, 8, 9, 3, 2. Even digits: 4, 8, 2 (3 total). Odd digits: 5, 9, 3 (3 total). Even digits must occupy even places (2nd, 4th, 6th). There are 3! ways to arrange even digits and 3! ways to arrange odd digits. Total = 3! * 3! = 6 * 6 = 36.

AI explanation

Using six digits means the six-digit number has three odd places (1st, 3rd, 5th) and three even places (2nd, 4th, 6th). The even digits are 4, 8, and 2, and they can be arranged in the three even places in 3 factorial ways, which equals 6. The remaining three odd digits (5, 9, 3) can be arranged in the three odd places in 3 factorial ways, which also equals 6. The total number of such natural numbers is 6 times 6, which is 36.

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