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Sagot :
The total number of ways to select 2 students from the 3 clubs is 55 ways.
Since there are three clubs which each has 10, 4 and 5 students, and we require the number of ways a teacher can select 2 students so that they are from different clubs.
Since order doesn't matter and anybody can be chosen first, we use combination theory.
Number of ways of selecting from the first two clubs
Since we have two slots, to select the first person from the first club, we ¹⁰C₁. For the second student from the club of 4, we have ⁴C₁. Also, there are 2 ways of selecting the two students.
So, there are ¹⁰C₁ × ⁴C₁/2!
= 10 × 4/2
= 20 ways from the first two clubs.
Number of ways of selecting from the next two clubs
For the next two clubs of 4 and 5 students, for the first slot, we have ⁴C₁. For the second student, we have ⁵C₁. Also, there are 2 ways of selecting the two students.
So, there are ⁴C₁ × ⁵C₁/2!
= 4 × 5/2
= 10 ways from the next two clubs.
Number of ways of selecting from the last two clubs
For the first and last club of 10 and 5 students, for the first slot, we have ¹⁰C₁. For the second student, we have ⁵C₁. Also, there are 2 ways of selecting the two students.
So, there are ¹⁰P₁ × ⁵P₁/2!
= 10 × 5/2
= 25 ways from the first and last club.
Total number of ways of selecting two students from the 3 clubs.
So, the total number of ways to select 2 students from the 3 clubs is 20 + 10 + 25 = 55 ways.
So, the total number of ways to select 2 students from the 3 clubs is 55 ways.
Learn more about combinations here:
https://brainly.com/question/25990169
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