To solve this problem, we need to calculate how many ways five students can be chosen from a class of 25 students without regard to the order. This is a combination problem, where the formula for combinations is:

$C(n, k) = \frac{n!}{k!(n-k)!}$

Where:

• $n$ is the total number of students (25 in this case),
• $k$ is the number of students to be selected (5 in this case),
• $!$ denotes factorial (the product of all positive integers up to that number).

So, we need to compute $C(25, 5)$:

$C(25, 5) = \frac{25!}{5!(25-5)!} = \frac{25!}{5! \cdot 20!}$

Since $25!$ and $20!$ share many factors, we can simplify this to:

$C(25, 5) = \frac{25 \times 24 \times 23 \times 22 \times 21}{5 \times 4 \times 3 \times 2 \times 1}$

Now, calculating the numerator and the denominator:

$25 \times 24 = 600$ $600 \times 23 = 13,800$ $13,800 \times 22 = 303,600$ $303,600 \times 21 = 6,375,600$

Now for the denominator:

$5 \times 4 \times 3 \times 2 \times 1 = 120$

Finally, dividing:

$\frac{6,375,600}{120} = 53,130$

Thus, there are 53,130 different ways to select five students from a group of 25.

Would you like further details on this, or do you have any other questions?

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Here are 5 related questions:

1. How many ways can 6 students be chosen from 30?
2. What is the difference between combinations and permutations?
3. How does the formula for combinations change if the order is important?
4. How can combinations be applied to probability problems?
5. What is a factorial and why is it important in combinatorics?

Tip: When calculating combinations, remember that order does not matter. This makes combinations different from permutations where order is important.