In one school, there are five teachers: teacher A, teacher B, teacher C, teacher D and teacher E. However, only three students (Mr X, Mr Y and Mr Z) have entered the school this year. Note that in this school, one teacher is supposed to teach one student per teacher. In this case, how many ways can the five teachers be allocated to these three students?
Question
In one school, there are five teachers: teacher A, teacher B, teacher C, teacher D and teacher E. However, only three students (Mr X, Mr Y and Mr Z) have entered the school this year. Note that in this school, one teacher is supposed to teach one student per teacher. In this case, how many ways can the five teachers be allocated to these three students?
Solution
This is a combination problem because the order in which the teachers are assigned to the students does not matter.
The formula for combinations is:
C(n, r) = n! / [r!(n-r)!]
where:
- n is the total number of options,
- r is the number of options chosen at a time,
- "!" denotes a factorial, which means multiplying all positive integers up to that number.
In this case, n = 5 (the total number of teachers) and r = 3 (the number of students).
So, the calculation would be:
C(5, 3) = 5! / [3!(5-3)!]
= 54321 / [321 * 2*1]
= 5*4 / 2
= 10
So, there are 10 ways the five teachers can be allocated to these three students.
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