How many combinations are possible? Assume the items are distinct.

7 items chosen 5 at a time

 combinations

To calculate the number of combinations when choosing \( r \) items from a set of \( n \) distinct items, we use the formula for combinations:

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

For this problem, you have \( n = 7 \) and \( r = 5 \). Plugging in these values:

\[
\binom{7}{5} = \frac{7!}{5!(7 - 5)!} = \frac{7!}{5! \times 2!}
\]

Calculating the factorials:

1. \( 7! = 5040 \)
2. \( 5! = 120 \)
3. \( 2! = 2 \)

Substitute these values into the formula:

\[
\binom{7}{5} = \frac{5040}{120 \times 2} = \frac{5040}{240} = 21
\]

### Answer
There are **21 possible combinations** when choosing 5 items from a set of 7 distinct items.

Would you like more details on the calculation process, or have any questions? 

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Here are 5 additional questions for exploration:

1. How many combinations are there if you choose 4 items from 7?
2. What if the items were identical? How would that affect the count?
3. How many ways can you arrange 5 items out of 7 (permutations)?
4. What if we change the number of items to 8? How does that affect combinations?
5. How many combinations are there for choosing 3 items out of 10?

#### Tip
When calculating combinations, remember that \( \binom{n}{r} = \binom{n}{n - r} \), simplifying work if \( r \) is more than half of \( n \).

How many combinations are possible? Assume the items are distinct. 7 items chosen 5 at a time.

Find the number of ways to choose 5 items from a set of 7 distinct items using the combination formula. This problem demonstrates combinatorial counting with step-by-step factorial calculations. Ideal for students in Grades 9-12.

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