Since either Maria or Nathan has to be part of the 5 member team,

Andrew can select 4 team members from the remaining 10 team members and add either Maria or Nathan as the 5th team members

To select 4 team members out of 10 = 10C4 = 10 ! / ( 6! * 4 !) = 210
Since either Maria or Nathan can be the 5th team member, total number of teams = 210 * 2 = 420

Answer is C. 420

We need to determine the number of ways Andrew can select a team with either Maria or Nathan.

If Maria is selected for the team but Nathan is not, the team can be selected in 10C4 ways (since Maria definitely makes the team and Nathan does not):

10C4 = (10 x 9 x 8 x 7)/4! = (10 x 9 x 8 x 7)/(4 x 3 x 2 x 1) = 10 x 3 x 7 = 210 ways

Since there are an additional 210 ways in which the team could be created (with Nathan and without Maria), Andrew can select the team in 210 + 210 = 420 ways.

Answer: C
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Hi,

Can you check please why I get wrong result? Here is how I calculate:
Favorable outcomes=Total outcomes-Unfavorable outcomes (when Maria and Nathan are together in the team)
Favorable outcomes=12!/7!5!-10!/3!7! (we have placed Maria and Nathan to the team of five and need to add other 3 people from the rest of 10 students) = 672

Thanks.

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:





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.

Your Favorable outcomes also include the scenario when Nathan & Maria both are not present. So you need to subtract that as well from Total outcomes.
Favorable outcomes=12!/7!5!-10!/3!7!-10!/5!5!(when both Nathan & Maria aren't present)=420

METHOD-1

Total Team of 5 out of 12 candidates = 12C5 = 792
Total Team of 5 out of 12 candidates without Maria and Nathan = 10C5 = 252
Total Team of 5 out of 12 candidates with Maria and Nathan both = 10C3 = 120

Favourable Outcomes = 792 - 252-120 = 420

Answer: Option C

METHOD-2

Total Team of 5 out of 12 candidates with Maria and without Nathan = 10C4 = 210
Total Team of 5 out of 12 candidates with Nathan and without Maria = 10C4 = 210

Favourable Outcomes = 210+210 = 420

Answer: Option C
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