Answer:
Suppose that you’ve chosen an odd digit, d, and two even digits, e0 and e1. If you use d only once, you can put it in any of 9 positions in the number. Then you can distribute e0 and e1 in the other 8 positions however you please, except that you cannot use just e0 or just e1. There are 28 ways to distribute e0 and e1 among the 8 open positions, but two of those use only one of the digits, so there are 28−2 ways that use both digits. Thus, we get a total of 9(28−2) nine-digit numbers in this subcase. If you use d twice, there are (92) pairs of positions in which you can place the two d’s, and by the same reasoning as before there are 27−2 ways to distribute e0’s and e1’s amongst the remaining 7 places, so there are (92)(27−2)
nine-digit numbers in this subcase. Continuing in this fashion, it’s a little tedious but not hard to get the number of nine-digit numbers in this case.
If instead you’ve chosen odd digits d0
and d1 and even digits e0,e1,e2, and e3, you can reason similarly, though there’s a little more work involved. Suppose, for instance, that you’ve decided to put the odd digits into 4 spots, leaving the other 5 for the even digits. There are (94) ways to choose the 4 positions for the odd digits. There are 24−2 ways to distribute d0 and d1 amongst these 4 positions, excluding the two ways that use only one of the two digits. It remains to count the ways to distribute the even digits amongst the other 5 positions. Clearly one of them must appear twice and the other three once each. There are 4 ways to pick the one that appears twice, and there are (52) ways to pick the slots for it. There are then 3! ways to permute the remaining 3 even digits amongst the remaining three slots, for a grand total of (94)(24−2)⋅4(52)⋅3! numbers. There a few more subcases, depending on how many places you fill with odd digits, but you can use the same basic ideas for all of them.
HOPE IT"S NOT TOO COMPLEX FOR YOU