The number of 9 digit numbers, that can be formed using all the digits of the number 123412341 so that the even digits occupy only even places, is ______________.

This problem involves permutations and combinations. Here are the steps to solve it:

1. First, identify the even and odd digits in the number 123412341. The even digits are 2, 4, 2, 4 and the odd digits are 1, 3, 1, 3, 1.

2. Next, identify the even and odd places in a 9-digit number. The even places are the 2nd, 4th, 6th, and 8th positions, and the odd places are the 1st, 3rd, 5th, 7th, and 9th positions.

3. The problem states that the even digits must occupy the even places. There are 4 even digits and 4 even places, so calculate the number of ways the even digits can be arranged in the even places. Since there are 2 each of the digits 2 and 4, the number of arrangements is 4! / (2! * 2!) = 6.

4. Similarly, calculate the number of ways the odd digits can be arranged in the odd places. There are 5 odd digits and 5 odd places. Since there are 3 of the digit 1 and 2 of the digit 3, the number of arrangements is 5! / (3! * 2!) = 10.

5. Finally, multiply the number of arrangements of even digits and odd digits to get the total number of 9-digit numbers. So, 6 * 10 = 60.

Therefore, the number of 9-digit numbers that can be formed using all the digits of the number 123412341 so that the even digits occupy only even places is 60.