You're exactly right about they're not being mutually exclusive.

This is the technique of counting by inclusion-exclusion.

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That's seems to be a demanding problem for a junior high school level algebra class. (Unless the teacher expects them to do it by hand out to 210?) Hey - if the students are up to it - fantastic!

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There's a trick you can use that's not ideal, in that it's exploiting that 210 is evenly divisible by 2 and 3 and 5. I'll show you how to do it the general correct way, but here's the trick:

Do it "by hand" for pumpkins #1-30 (You'll get Good are 1, 7, 11, 13, 17, 19, 23, 29, for a total of 8). Notice that "every 2nd", "every 3rd", and "every 5th" all stop on pumpkin #30, and then begin to repeat their pattern exactly as if you started counting at 1 again. In other words, the number of bad pumpkins from 1-30 is the same as the number of bad pumkins from 31-60. And also for 61-90, and 91-120, and 121 - 150, and 151-180, and 181-210. Thus the number of Good pumpkins between 1 and 210 is 7 times the number of good pumpkins between 1 and 30. (So the number of Good pumpkins of those 210 is 7x8 = 56.)

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Here's the proper way to do it:

Let X = {1, 2, 3, ... , 210}

Let M2 = the set of multiples of 2 in X = {2, 4, 6, ..., 210}

Let M3 = the set of multiples of 3 in X = {3, 6, 9, ..., 210}

Let M5 = the set of multiples of 5 in X = {5, 10, 15, ..., 210}

Then Bad = |M2 union M3 union M5| (the notation |Z| means the number of things in set Z).

Here's how inclusion-exclusion counting works:

You can count each of those sets by dividing into 210. All of those numbers (2, 3, 5, 6, 10, 15, 30) are divisors of 210, so you don't even have the complications from rounding down.

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Counting by inclusion-exclusion:

That's because anything that's in the intersection was counted twice in the initial , so you has to subtract it once so that it gets counted a net of only 1 time.

If in a room, 5 people raise their hands to say they like football, 3 people raise their hands to say they like basketball, and exactly 2 people raised their hands both times, then how many people in that room like football or basketball? Try 5 + 3 = 8, but then remember that 2 people raised their hands both times, and so those two people were double counted, so the actual correct answer is 8-2 = 6.