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Pick any $2$ of the $12$ people to make the $2$ person team, and then any $5$ of the remaining $10$ for the first team of $5,$ and then the remaining $5$ are on the other team of $5;$ this overcounts by a factor of $2$ though, since there is no designated “first” team of $5.$ So the number of possibilities is  $\left(\left(^{12}\text{C}_{2}\right)\left(^{10}\text{C}_{5}\right)\right)/2 = 8316.$
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