How many number can be formed with digits 1,2,3,4,3,2,1 so that odd digits always occupy the odd places?a)22b)18c)32d)14Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Answer

Approach:
To solve this problem, we need to determine the number of ways we can arrange the digits 1, 2, 3, 4, 3, 2, 1 such that odd digits always occupy the odd places.

Counting Odd Digits:
- There are three odd digits in the set: 1, 3, 3.
- There are 3! = 6 ways to arrange these odd digits in the odd places.

Counting Even Digits:
- There are two even digits in the set: 2, 4.
- There are 2! = 2 ways to arrange these even digits in the even places.

Calculating Total Number of Ways:
- Multiply the number of ways to arrange odd digits and even digits: 6 * 2 = 12.
- Since the given set has repeated digits, we need to consider the number of ways to arrange these repetitions.
- The repeated digits 2 and 3 can be arranged in 2! = 2 ways.
- Therefore, the total number of ways = 12 * 2 = 24.

Answer:
Therefore, the correct answer is 24, which is closest to option B (18).

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