How many unique six-digit odd numbers can be formed by the digits $1, 1, 2, 3, 3$, and $3$?

Question

This is the 14th question on the 2018 Blaine WSMC. I tried but just can't figure out the answer since probability is my weak spot, can someone help me? Thank you.

Answer 1

Either 1 is the units digit or 3 is. If 1 is the units digit, I can place the other 1 in any of the five remaining places, then the 2 in the remaining four, after which the rest must be filled with 3's. This gives $5×4=20$ ways.

If 3 is the units digit, I can place the 1's in $\binom52=10$ ways and the 2 in 3 ways after that, after which the rest must be filled with 3's. This gives $10×3=30$ ways.

Adding the two sub-results gives the final answer of 50 numbers.

Answer 2

Alternatively: There are $6!=720$ permutations of the digits, not taking account of the fact that some of them are identical. We must divide by $2!=2$ because swapping the 1s makes no difference, and divide by $3!=6$ because permuting the 3s makes no difference. This yields $720/2/6=60$. But we must count only odd numbers. One of these numbers is odd iff the 2 is not in the last position. The 2 appears in each of the 6 positions in $60/6=10$ of these numbers, so 10 of these numbers are even. We must exclude those 10, leaving $60-10=50$ even numbers.

Answer 3

You must exclude the uniqueness condition, since it is meaningless.

You've got $\binom6{2,1,3}=60$ different such numbers without the oddness condition. Of those, $1$ in $6$ has the $2$ at the end and is therefore even (you can group the numbers by cyclic permutations if you want proof). So there are $\frac56\times60=50$ odd such numbers (odd positive integers whose decimal representation has two occurrences of the digit$~1$, one occurrence of the digit$~2$, three occurrences of the digit$~3$, and no other digits).