I like your approach. When I solved it, I did it a bit differently.

There are 3 spots

Spot #1 can be any of the 10, because no one has been selected so there is no one to avoid.

Spot #2 can choose 1 of only 8 people. 9 people left, but we cannot select the spouse of #1

Spot #3 can choose 1 of only 6 people. 8 people left but we can't select the spouse of #1 or spouse of #2.

Now, we have 8 x 10 x 6 = 480 total possibilities, but we're not done.

This gives us the Permutation of the event, where we have identified order of #1, #2, #3, etc but the order doesn't matter. So we need to divide out the # of times (or permutations) of when we have the same people but in a different order. The total # of perms of a group of 3 is 3x2 = 6

If you remember we had 80 * 6, and we're dividing out 6, this leaves 80.

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J Allen Morris

**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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