ExplanationThis is a classic combinatorics problem in which order doesn’t matter—that is, picking Javier and Sonya is the same as picking Sonya and Javier. A person is either on the team or not. Use the standard “order doesn’t matter” formula:

For Quantity A:\(\frac{6!}{2! \times 4!}\)

For Quantity B:

\(\frac{6!}{4! \times 2!}\)

The two quantities are equal. Note that it is not actually necessary to reduce each quantity. The factorials are the same in each, so the resulting quantities must be equal.

This will always work—when order doesn’t matter, the number of ways to pick 4 and leave out 2 is the same as the number of ways to pick 2 and leave out 4. Either way, it’s one group of 4 and one group of 2. What actually happens to those groups (getting picked, not getting picked, getting a prize, losing a contest, etc.) is irrelevant to the ultimate solution.

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Sandy

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