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 ______

We need to find the number of 9-digit numbers formed using all digits of 123412341, where even digits occupy only even places.

We start by identifying the digits in the number 123412341. The digits are: 1, 2, 3, 4, 1, 2, 3, 4, 1.

Here, 1 appears 3 times, 2 appears 2 times, 3 appears 2 times, and 4 appears 2 times. Thus, the even digits are 2, 2, 4, 4 (4 even digits) and the odd digits are 1, 1, 1, 3, 3 (5 odd digits).

Next, we note that in a 9-digit number, the even places are positions 2, 4, 6, and 8. Since there are exactly 4 even digits and 4 even places, all even digits must occupy the even positions, and the odd digits occupy the odd positions.

We then arrange the even digits 2, 2, 4, 4 in the 4 even places. The number of arrangements is $$\frac{4!}{2! \times 2!} = 6$$.

Next, we arrange the odd digits 1, 1, 1, 3, 3 in the 5 odd places (positions 1, 3, 5, 7, 9). The number of arrangements is $$\frac{5!}{3! \times 2!} = 10$$.

Multiplying these results gives the total number of such 9-digit numbers as $$6 \times 10 = 60$$.

Therefore, the total number of 9-digit numbers that can be formed under the given conditions is 60.