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GMAT Quant — Probability & Combinations

A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?

Explanation	Calculations	Help
$We first calculate the number of ways to choose one mathematics candidate from seven eligible candidates.$	$\frac{7 !}{1 ! 6 !} = \frac{7 \times 6 !}{1 \times 6 !} = 7$	Theory & Tactics Method Card PC3-C Click to view full details
$We then calculate the number of ways to choose two computer science candidates from ten eligible candidates, where the two positions are identical.$	$\frac{10 !}{2 ! 8 !} = \frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !} = \frac{10 \times 9}{2 \times 1} = 45$	Theory & Tactics Method Card PC3-C Click to view full details
$Because the mathematics selection and the computer science selection are independent, we multiply these counts to find the total number of sets of three candidates.$	$7 \times 45 = 315$	Theory & Tactics Method Card PC2-A Click to view full details

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315

GMAT Quant — Probability & Combinations

A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?

Explanation	Calculations	Help
$We first calculate the number of ways to choose one mathematics candidate from seven eligible candidates.$	$\frac{7 !}{1 ! 6 !} = \frac{7 \times 6 !}{1 \times 6 !} = 7$	Theory & Tactics Method Card PC3-C Click to view full details
$We then calculate the number of ways to choose two computer science candidates from ten eligible candidates, where the two positions are identical.$	$\frac{10 !}{2 ! 8 !} = \frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !} = \frac{10 \times 9}{2 \times 1} = 45$	Theory & Tactics Method Card PC3-C Click to view full details
$Because the mathematics selection and the computer science selection are independent, we multiply these counts to find the total number of sets of three candidates.$	$7 \times 45 = 315$	Theory & Tactics Method Card PC2-A Click to view full details

Scroll horizontally to view all columns

315