To solve this problem, let's consider each scenario separately:

1. When the Order Is Taken into Consideration

Since the order matters, we are dealing with permutations. We want to choose 3 objects out of 4, and we care about the sequence in which they are chosen.

The number of ways to do this is given by the permutation formula: $P(n, r) = \frac{n!}{(n - r)!}$ where $n = 4$ (the total objects) and $r = 3$ (the number of objects chosen).

Calculating: $P(4, 3) = \frac{4!}{(4 - 3)!} = \frac{4!}{1!} = \frac{24}{1} = 24$

So, there are 24 ways to choose 3 objects from 4 if the order matters.

2. When the Order Is Not Taken into Consideration

If the order doesn't matter, we're dealing with combinations. We want to choose 3 objects out of 4 without regard to sequence.

The number of ways to do this is given by the combination formula: $C(n, r) = \frac{n!}{r!(n - r)!}$

Substituting the values: $C(4, 3) = \frac{4!}{3!(4 - 3)!} = \frac{4!}{3! \cdot 1!} = \frac{24}{6 \cdot 1} = 4$

Thus, there are 4 ways to choose 3 objects from 4 if the order does not matter.

Summary

• With order consideration (permutations): 24 ways
• Without order consideration (combinations): 4 ways

Would you like more details or have any questions?

---

Here are some related questions to consider:

1. How would the calculations change if we wanted to choose 2 objects instead of 3?
2. What if we had 5 objects to choose from instead of 4?
3. How does the formula for permutations differ if we are allowed to replace objects after choosing?
4. How would we calculate the number of ways to choose all objects without considering the order?
5. Can you think of a real-life scenario where combinations or permutations would be useful?

Tip: When the problem states that "order matters," think about permutations; when it says "order does not matter," think about combinations.