Too many wedding rings.

For 0 rings, there is one way.

For 1 ring, there are three ways to choose the ring, then five ways to choose the finger, so $3 \cdot 5 = 15$ ways.

For 2 rings, there are three ways to choose the two rings. If both rings are on different fingers, there are $5 \cdot 4 = 20$ ways. If both rings are on the same finger, there are $5 \cdot 2 = 10$ ways. This gives $3 \cdot (20 + 10) = 90$ ways in this case.

For 3 rings, either (i) all rings are on different fingers, (ii) two rings are on one finger and one ring is on a different finger, or (iii) all rings are on the same finger. There are $5 \cdot 4 \cdot 3 = 60$ , $3 \cdot 5 \cdot 4 \cdot 2 = 120$ , $5 \cdot 6 = 30$ ways, respectively, for a total of $60 + 120 + 30 = 210$ ways in this case.

The total number of ways is $1 + 15 + 90 + 210 = 316$ .

Case 1: One ring. There are $3$ options for which ring you wear, and $5$ options for which finger to put it on, so $5\times3$ .

Case 2a: Two rings, each on separate fingers. There are $\binom{3}{2}$ ways to choose the two rings and $P(5,2)=5\times4$ ways to choose which fingers to put them on, so $\binom{3}{2}\times5\times4$ .

Case 2b: Two rings, both on a single finger. There are $\binom{3}{2}$ ways to choose the two rings, $5$ ways to choose the finger, and $2!$ ways to stack the rings, so $\binom{3}{2}\times5\times2!$ .

Case 3a: Three rings, each on separate fingers. There are $P(5,3)=5\times4\times3$ ways to put the rings on separate fingers.

Case 3b: Three rings, two are stacked and one is alone. There are $\binom{3}{2}$ ways to choose the rings to be stacked, $P(5,2)=5\times4$ ways to choose the fingers (the stacked ring acts like a single ring), and $2!$ ways to arrange the stacked ring, so $\binom{3}{2}\times5\times4\times2!$

Case 3c: Three rings, all stacked together. There are $5$ options for which finger on which to place all three rings, and once you choose, there are $3!$ ways to stack the rings, so $5\times3!$

Case 4: No rings. Only $1$ way to do this.

Total: $5\times3+P(5,2)\times\binom{3}{2}+5\times\binom{3}{2}\times2!+P(5,3)+5\times4\times\binom{3}{2}\times2!+5\times3!+1=\boxed{316}$

Cases will be based on the number of rings used.

Case 1 (No rings): 1
Case 2 (1 ring): 5 * 4
Case 3 (2 rings):
Placement 1 (2 rings on same finger): 5 * (4 choose 2) * 2!
Placement 2 (2 rings on different fingers): (5 choose 2) * 4 * 3
Case 4 (3 rings): * Here comes the hard part Placement 1 (one ring on each finger): (5 choose 3) * 4 * 3 * 2
Placement 2 (two rings on one finger and one ring on the other): 5 * 4 * (4 choose 2) * 2!
Placement 3 (three rings on the same finger): 5 * (4 choose 3) * 3!

1 + 20 + 60 + 120 + 240 + 240 + 120 = 801 ways