Help on probability problem

Kindly edit your post to have only 1 question per post. Thank you

out of 7, 3 digits gone.
so, 4 left
we want 3 odd numbers out of 4 digits

answer will be 3P3 /4P3

= 3.2 / 4.3.2
=
1/4

What is 3P3? How do you meet condition that there can be no 1 digit?

I tried to use combinations with repetitions, could you help me why it doesn't work?

4 digits, 3 of them should be odd, 1 even. Odd can be from set: 3, 5, 7, 9; even can be from set: 2, 4, 6, 8.

If C~_n^k is a number of combinations with repetitions from n number in k places, then

total = C~_8^4 (because 8 possible digits can be, 4 places)
odd = C~_4^3 (because 4 odd can be, 3 places)
even = C_4^1 (or 4 variants - 2, 4, 6, 8).

Also, C~n_k = C_(n+k-1)^k)

So formula is:
odd*even/total = (C_6^3*C_4*1)/(C_11^4)=8/33

Which is close to 1/4 though Where's the error?