Before you even get to the restriction on this problem, first identify that it is a combination problem - the order does not matter, but rather it's just about who's on the team. So prepare to use the Combinations formula: N!/K!(N−K!)
Whenever you encounter a combinations problem with restrictions, you have two major options:
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1) Solve for the number of unrestricted combinations, then find and subtract the number of restricted combinations. Here the restricted groups are "neither Nathan nor Maria is on the team" and "both Nathan and Maria are on the same team."
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2) Solve directly for the combinations that do not violate the restriction. Here that means considering two case: Maria is on the team and Nathan is not, or Nathan is on the team and Maria is not.
For this problem, note that the calculation will be the same for both options in #2: in each case, one spot is already "spoken for," one person is no longer eligible to be part of the pool of N candidates, and 10 people are left for the 4 remaining spots. Therefore it's best on this problem to proceed by solving directly for the combinations that do not violate the restriction.
That calculation will look like:
Maria is on the team, Nathan is not: 10 candidates for 4 remaining spots = 10!/4!6!=(10×9×8×7)/(4×3×2×1)
Nathan is on the team, Maria is not: the math is the same, with 10 candidates for 4 remaining spots. This is also 210
Therefore the correct answer is the sum of those two totals, 210+210=420
Choice C is correct.