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13 Feb 2020, 07:27
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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76% (01:51) correct
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There are 5 students and 3 teachers. In how many ways can a team of 5 be formed so that there is at least one teacher but not more than two teachers in the team?
A. 450
B. 180
C. 60
D. 45
E. 30
A. 450
B. 180
C. 60
D. 45
E. 30
13 Feb 2020, 08:54
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We have two ways to make a suitable team, so we have two cases, and we'll need to add the numbers from each case. We can either:
- use 1 teacher and 4 students. Then we have 3 choices for the teacher, and 5 choices for the students (since we're just leaving one student out), and, multiplying our choices, we have 3*5 = 15 possible teams
- use 2 teachers and 3 students. Then we have 3 choices for the teachers (since we're just picking which teacher to leave out), and 5C3 = (5)(4)/2 = 10 choices for the two students, since we're picking 2 from 5 and order does not matter. Multiplying our choices, we can make 3*10 = 30 such teams.
Adding the results from each case, we have 15+30 = 45 possible teams in total.
- use 1 teacher and 4 students. Then we have 3 choices for the teacher, and 5 choices for the students (since we're just leaving one student out), and, multiplying our choices, we have 3*5 = 15 possible teams
- use 2 teachers and 3 students. Then we have 3 choices for the teachers (since we're just picking which teacher to leave out), and 5C3 = (5)(4)/2 = 10 choices for the two students, since we're picking 2 from 5 and order does not matter. Multiplying our choices, we can make 3*10 = 30 such teams.
Adding the results from each case, we have 15+30 = 45 possible teams in total.
13 Feb 2020, 13:43
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students = 5
teachers = 3
total = 8
team capacity = 5
condition = teacher must be (atleast 1 and not more than 2)
possible ways : T S S S S OR T T S S S = 3C1*5C4 + 3C2*5C3 => 3*5+3*10 => 45 ways//
teachers = 3
total = 8
team capacity = 5
condition = teacher must be (atleast 1 and not more than 2)
possible ways : T S S S S OR T T S S S = 3C1*5C4 + 3C2*5C3 => 3*5+3*10 => 45 ways//
