Final answer:
By arranging the odd digits in the even places and calculating permutations, we conclude that there are 60 possible numbers that can be formed, which does not align with any of the provided multiple-choice answers, suggesting an error in the options or the question itself.
Step-by-step explanation:
To find the numbers formed by the digits 1, 1, 2, 2, 2, 2, 3, 4, and 4 with odd digits in even places, we'll start by assigning places to the odd and even digits. There are 5 even places in a 9-digit number. Since we have three odd digits (1, 1, and 3), the locations for these can be chosen in 5 choose 3 ways, which is 10 ways.
For each of these arrangements, the odd digits can be permuted in 3!/2! = 3 ways, because we have two 1s. The remaining places are to be filled with even numbers 2, 2, 2, 2, and 4, 4. These can be arranged in 6! / (4!2!) = 15 ways. Now multiplying, we get 10 (ways to choose places) × 3 (ways to permute odd digits) × 15 (ways to arrange even digits), resulting in 450 possible numbers.
However, we also have to consider that every even place must be filled with an odd number, which limits our permutations. For this case, we have to divide the total by the number of even places which is 5, thus we get 450 / 5 = 90. Finally, since the question requires numbers with the odd digits on even places only, we consider that the starting even place can have only two options (1 or 3), not three. This further divides our number of possibilities by 3/2, giving us 90 × 2/3 = 60 possible numbers, which isn't an option in the original multiple-choice answers provided. Therefore, there seems to be a mistake in the choices or the framing of the question.