How many possible passcode in 6-digit passcode that contains exactly four odd digits (can't be repeated)

Question

Where each element can be any digit from 0 to 9.

Checking if I did it correctly,

$5*4*5*4*3*2$ = 2,400

5, 4 being evens

5, 4, 3, 2 being odd digits

Total # = $15*2,400$ = 36,000 possibilities.

Answer 1

Another way to look at it:

Pick your two different even digits ($_5C_2 = 10$), pick which two spaces they go in ($_6C_2 = 15$) and which order they appear in ($2$).

Then arrange four odd digits in the spaces that remain ($5 \cdot 4 \cdot 3 \cdot 2 = 120$).

This gives $36000$ codes.