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One way is to count by 2's five times: 2, 4, 6, 8, 10

The other way is to count by 5's two times: 5, 10

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Q: What are 2 ways to skip count to find 2 times 5?
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How do you find the sum of the interior of a dodecagon?

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How many unique ways can thirteen be obtained from the sum of 5 odd numbers?

I count 5, not counting the different orders that they are added as different ways. Here they are. If I missed some, then somebody else can add to it:3 + 3 + 3 + 3 + 1 = 133 + 3 + 5 + 1 + 1 = 135 + 5 + 1 + 1 + 17 + 3 + 1 + 1 + 19 + 1 + 1 + 1 + 1If you want to count different orders as unique, For #1, there are 4 additional ways (I wouldn't say that the different 3's are unique, so the 1 in each position).For #2, I count 13 additional ways. For # 3, I count 8 additional. For #4, I count 7 additional ways. And for #5, I count 4 additional ways. So that would be 36 additional ways, rearranging the orders, for a total of 41 ways.


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By ones By twos By fours By eights By tens By twenties By fourty (if you want to count that one)


How to solve probability questions?

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