Eleven students put their names on slips of paper inside a box. Three names are going to be taken out. How many different ways can the three names be chosen?

This problem is a combination problem where you need to determine the number of ways to choose 3 names out of 11 students.

The formula for combinations is given by:

\[
\text{Number of combinations} = \binom{n}{r} = \frac{n!}{r!(n-r)!}
\]

Where:
- \( n \) is the total number of items to choose from (in this case, 11 students).
- \( r \) is the number of items to choose (in this case, 3 names).

Substituting the values:

\[
\binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165
\]

So, there are **165 different ways** to choose 3 names out of 11 students.

Would you like further details or have any other questions?

Here are five related questions:
1. How many ways can you choose 4 names from 11 students?
2. What if there were 15 students instead of 11—how many ways could you choose 3 names then?
3. How does the number of combinations change if you need to choose 2 names instead of 3?
4. If the order of selection mattered, how would you calculate the number of arrangements?
5. How many different ways can 11 students be divided into groups of 3?

**Tip:** When dealing with combinations, remember that the order of selection does not matter—only the items chosen do.

Discover how many different ways you can choose 3 names from a group of 11 students with this combinatorics problem. Learn the combinations formula and step-by-step solution. Suitable for Grades 6-8.

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