Problem:

There are 12 students in a classroom, and the teacher needs to select a group of 3 students to represent the class in a competition. How many different ways can the teacher select the group of 3 students?

Solution:

To solve this, we use the formula for combinations because the order in which the students are selected does not matter. The formula for combinations is:

$\text{Combination formula:} \binom{n}{r} = \frac{n!}{r!(n-r)!}$

Here, $n = 12$ (total number of students) and $r = 3$ (students to be selected). Applying the combination formula:

$\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!}$

Now, let's break down the factorials:

• $12!$ (12 factorial) is the product of all integers from 12 down to 1, but since $9!$ is in the denominator, we can cancel out most terms and simplify the expression.

$\frac{12!}{3!9!} = \frac{12 \times 11 \times 10 \times 9!}{3! \times 9!} = \frac{12 \times 11 \times 10}{3!}$

Now calculate $3!$:

$3! = 3 \times 2 \times 1 = 6$

So the expression becomes:

$\frac{12 \times 11 \times 10}{6}$

Now calculate the product and division:

$12 \times 11 = 132$ $132 \times 10 = 1320$ $\frac{1320}{6} = 220$

Final Answer:

There are 220 different ways to select 3 students from a group of 12.

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Would you like more details or have any questions?

Related questions:

1. How would the problem change if you had to select 4 students instead of 3?
2. Can you use this combination formula to calculate how many ways to select 2 students from 12?
3. What would happen if the order of selection mattered? How would that change the formula?
4. How can you apply the same formula if you wanted to select no students (r = 0)?
5. Can you find a general pattern for selecting groups of different sizes from a given number of students?

Tip:

Whenever the order of selection doesn't matter, always use combinations (nCr). If the order does matter, use permutations!