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 $n$.