PyjamaScientist wrote:
If the question went like this- There are 6 couples in a room from which 4 people are selected. There are how many ways that there is exactly one couple selected? How would you have done this using your method?
You would then have 6 choices for the couple. Once you've chosen the couple, you have 10 choices for the third person on the team, and 8 choices for the fourth person, but the order of those two people doesn't matter, so we'd divide by 2!. So we'd have 6 * (10*8/2!) = 240 options in total.
You can confirm that answer in a different way. From the solutions earlier in the thread, we see there are (coincidentally) also 240 ways to choose no couples. If we wanted to choose two couples, we'd have 6 choices for the first, and 5 for the second, but since their order wouldn't matter we'd have 6*5/2! = 15 ways to choose two couples. Since we can't possibly choose more than two couples if we're only picking 4 people, adding the number of ways to choose zero, one and two couples, we find there should be 240 + 240 + 15 = 495 options altogether if we had no restrictions at all on how many couples we choose. But if we have no restrictions at all, we're just picking 4 people from 12, so we have 12C4 = (12)(11)(10)(9)/4! = 11*5*9 = 495 options. So that confirms the answer is correct, and also offers an alternative (but longer) way to solve the problem; to answer your question, you could count all of the possibilities with no restrictions, then subtract the options you don't want to count (zero couples and two couples).
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