Key Concepts and Formulas

• Permutations with Repetitions: The number of distinct permutations of $n$ objects, where $n_1$ are of one kind, $n_2$ are of another kind, ..., $n_k$ are of the $k$-th kind, is given by $\frac{n!}{n_1! n_2! \dots n_k!}$.
• Combinations: The number of ways to choose $k$ objects from a set of $n$ distinct objects is given by ${n \choose k} = \frac{n!}{k!(n-k)!}$.

Step-by-Step Solution

We are given the digits 1, 1, 2, 2, 2, 2, 3, 4, 4. We need to find the number of 9-digit numbers formed using these digits such that the odd digits occupy even places.

Step 1: Identify Odd and Even Digits and Their Counts

• Odd digits: 1 (appears twice), 3 (appears once). Total: 3 odd digits.
• Even digits: 2 (appears four times), 4 (appears twice). Total: 6 even digits.
• Total number of digits: 9.

Step 2: Identify Even and Odd Places

• Total places: 9 (1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th).
• Even places: 4 (2nd, 4th, 6th, 8th).
• Odd places: 5 (1st, 3rd, 5th, 7th, 9th).

Step 3: Place Odd Digits in Even Places

We need to place the 3 odd digits (1, 1, 3) in the 4 even places.

1. Choose 3 even places out of 4: We need to select 3 places out of the 4 even places for the odd digits. The number of ways to do this is given by the combination formula: ${4 \choose 3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4$ We use combinations because the order in which we choose the places does not matter.
2. Arrange the odd digits (1, 1, 3) in the chosen 3 places: Since the digit 1 is repeated twice, the number of distinct arrangements is given by the permutation formula with repetitions: $\frac{3!}{2!1!} = \frac{3 \times 2 \times 1}{(2 \times 1)(1)} = 3$
3. Total ways to place the odd digits: Multiply the number of ways to choose the places by the number of ways to arrange the digits in those places: $4 \times 3 = 12$

Step 4: Place Even Digits in Remaining Places

After placing the 3 odd digits, we are left with 6 even digits (2, 2, 2, 2, 4, 4) and 6 remaining places.

1. Arrange the 6 even digits (2, 2, 2, 2, 4, 4) in the remaining 6 places: The number of ways to arrange these digits is given by the permutation formula with repetitions: $\frac{6!}{4!2!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1)(2 \times 1)} = \frac{720}{24 \times 2} = \frac{720}{48} = 15$

Step 5: Calculate Total Arrangements

The total number of arrangements is the product of the number of ways to place the odd digits and the number of ways to place the even digits: $12 \times 15 = 180$

Common Mistakes & Tips

• Remember to divide by the factorials of the counts of repeated digits when calculating permutations.
• Distinguish between combinations (choosing places) and permutations (arranging digits).
• Always address the constraints first.

Summary

By first placing the odd digits in the even places and then arranging the remaining even digits in the remaining places, we found the total number of such numbers to be 180.

Final Answer The final answer is \boxed{180}, which corresponds to option (D).