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At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club. If each club has more than 5 students, how many such teams can be formed in which at least one student belongs to both clubs?
(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.
(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.
07 Aug 2025, 07:15
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Official Solution:
At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club. If each club has more than 5 students, how many such teams can be formed in which at least one student belongs to both clubs?
Let:
• \(x =\) number of students in only the math club
• \(y =\) number of students in only the science club
• \(z =\) number of students in both clubs
Total number of 2-person project teams where one student is from the math club and the other from the science club is \(xy + z(x + y) + zC2\). Where:
• \(xy =\) both students are in only one club each (math-only * science-only)
• \(z(x + y) =\) one student from both clubs and the other from only one club
• \(zC2 =\) both students are from both clubs We are asked: how many such teams have at least one student who is in both clubs?
So we want: \(z(x + y) + zC2\)
(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
This implies \(xy = 9\), which allows multiple \((x, y)\) pairs such as (1, 9), (3, 3), or (9, 1). In addition, we have no information about \(z\), so this is not sufficient.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.
This implies \(zC2 = 6\), which gives \(\frac{z!}{2!(z - 2)!} = 6\), or \(\frac{z(z - 1)}{2} = 6\). Solving gives \(z = 4\). But we don’t know \(x\) or \(y\), so not sufficient.
(1)+(2) From (1), we found that \((x, y)\) can be (1, 9), (3, 3), or (9, 1), and from (2), we know that \(z = 4\). However, if \((x, y)\) is (1, 9) or (9, 1), then the number of members in one of the clubs, math or science, becomes \(1 + z = 5\), which violates the condition that each club has more than 5 students. So \((x, y)\) must be (3, 3). We know the values of each unknown, and thus we can answer the question. Sufficient.
Answer: C
At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club. If each club has more than 5 students, how many such teams can be formed in which at least one student belongs to both clubs?
Let:
• \(x =\) number of students in only the math club
• \(y =\) number of students in only the science club
• \(z =\) number of students in both clubs
Total number of 2-person project teams where one student is from the math club and the other from the science club is \(xy + z(x + y) + zC2\). Where:
• \(xy =\) both students are in only one club each (math-only * science-only)
• \(z(x + y) =\) one student from both clubs and the other from only one club
• \(zC2 =\) both students are from both clubs
So we want: \(z(x + y) + zC2\)
(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
This implies \(xy = 9\), which allows multiple \((x, y)\) pairs such as (1, 9), (3, 3), or (9, 1). In addition, we have no information about \(z\), so this is not sufficient.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.
This implies \(zC2 = 6\), which gives \(\frac{z!}{2!(z - 2)!} = 6\), or \(\frac{z(z - 1)}{2} = 6\). Solving gives \(z = 4\). But we don’t know \(x\) or \(y\), so not sufficient.
(1)+(2) From (1), we found that \((x, y)\) can be (1, 9), (3, 3), or (9, 1), and from (2), we know that \(z = 4\). However, if \((x, y)\) is (1, 9) or (9, 1), then the number of members in one of the clubs, math or science, becomes \(1 + z = 5\), which violates the condition that each club has more than 5 students. So \((x, y)\) must be (3, 3). We know the values of each unknown, and thus we can answer the question. Sufficient.
Answer: C
15 Sep 2025, 21:30
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I like the solution - it’s helpful.
