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The expression \( 7C_5 \) is a combination, also known as "7 choose 5." It calculates how many ways you can choose 5 elements from a set of 7. The formula for combinations is:

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

For \( 7C_5 \), we substitute \( n = 7 \) and \( r = 5 \):

\[
7C_5 = \frac{7!}{5!(7 - 5)!} = \frac{7!}{5!2!}
\]

Now, let's calculate this step-by-step:

1. \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \)
2. \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
3. \( 2! = 2 \times 1 = 2 \)

Now, plug in these values:

\[
7C_5 = \frac{5040}{120 \times 2} = \frac{5040}{240} = 21
\]

Thus, \( 7C_5 = 21 \).

Would you like more details on combinations or have any other questions?

Here are five related questions to explore:
1. What is the difference between permutations and combinations?
2. How would you calculate \( 8C_3 \)?
3. Can you explain the general use of combinations in probability theory?
4. What is the factorial of a number and how is it used in combinations?
5. How would you solve a permutation problem like \( 7P_5 \)?

**Tip:** Remember that combinations don't care about the order of selection, unlike permutations, where order matters.

Learn how to evaluate the combination 7C5 using the formula nCr = n! / (r!(n - r)!). This problem involves basic combinatorics and factorials. Suitable for students in Grades 9-12.

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