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.