Question

Suppose you have a group of 12 students and want to create study groups of five in which order doesn’t matter. How many different combinations are there?


385


7!


1140


792

Answer 1

To find the number of different combinations of 12 students taken 5 at a time, you can use the combinations formula:

\[
C(n, r) = \frac{n!}{r!(n-r)!}
\]

Where:
- \( n \) is the total number of items (students) to choose from,
- \( r \) is the number of items (students) to choose,
- \( ! \) denotes factorial, which is the product of all positive integers up to that number.

In this case, \( n = 12 \) and \( r = 5 \):

\[
C(12, 5) = \frac{12!}{5!(12-5)!} = \frac{12!}{5! \cdot 7!}
\]

Calculating the factorials needed:

- \( 12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7! \) (we can cancel \( 7! \) in the numerator and denominator)
- \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

Now substituting these into the formula:

\[
C(12, 5) = \frac{12 \times 11 \times 10 \times 9 \times 8}{5!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{120}
\]

Calculating the numerator:

\[
12 \times 11 = 132
\]
\[
132 \times 10 = 1320
\]
\[
1320 \times 9 = 11880
\]
\[
11880 \times 8 = 95040
\]

Now, divide by \( 120 \):

\[
\frac{95040}{120} = 792
\]

Thus, the number of different combinations of 12 students taken 5 at a time is **792**.

So the correct answer is **792**.