Math worksheet on parking lot arrangements with colored cars and logic puzzles.

Let's solve this step by step.

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Problem Summary:

We have:
- 4 cars: Red, Blue, Green, Yellow
- 4 parking lots: Lot 1, Lot 2, Lot 3, Lot 4
- Constraint: Lot 1 can only be used by the Red or Yellow car.
- All other lots (2, 3, 4) can take any car.
- Each car must go into a different lot (no two cars in the same lot).
- We are told there are 12 possible ways to park the cars under these rules.
- We need to find all 12 valid arrangements, and then answer the second question.

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Step 1: Understand the Constraint

Lot 1 → only Red or Yellow car allowed.

So, we split our cases based on what is in Lot 1.

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Case 1: Red car is in Lot 1

Then:
- Lot 1 = Red
- Remaining cars: Yellow, Blue, Green
- Remaining lots: 2, 3, 4

We now assign the remaining 3 cars to the 3 lots. That’s just a permutation of 3 items:

Number of permutations = 3! = 6

Let’s list them:

| Lot 1 | Lot 2 | Lot 3 | Lot 4 |
|-------|-------|-------|-------|
| Red | Yellow| Blue | Green |
| Red | Yellow| Green | Blue |
| Red | Blue | Yellow| Green |
| Red | Blue | Green | Yellow|
| Red | Green | Yellow| Blue |
| Red | Green | Blue | Yellow|

✔ So, 6 arrangements when Red is in Lot 1.

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Case 2: Yellow car is in Lot 1

Then:
- Lot 1 = Yellow
- Remaining cars: Red, Blue, Green
- Remaining lots: 2, 3, 4

Again, 3! = 6 ways to arrange the remaining cars.

List them:

| Lot 1 | Lot 2 | Lot 3 | Lot 4 |
|-------|-------|-------|-------|
| Yellow| Red | Blue | Green |
| Yellow| Red | Green | Blue |
| Yellow| Blue | Red | Green |
| Yellow| Blue | Green | Red |
| Yellow| Green | Red | Blue |
| Yellow| Green | Blue | Red |

✔ Another 6 arrangements when Yellow is in Lot 1.

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✔ Total Valid Arrangements:

6 (Red in Lot 1) + 6 (Yellow in Lot 1) = 12

This matches the hint in the problem!

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✔ Now fill in the table:

We already listed 12 possibilities. Let's write them in the table format.

| LOT 1 | LOT 2 | LOT 3 | LOT 4 |
|-------|-------|-------|-------|
| RED | YELLOW| BLUE | GREEN |
| RED | YELLOW| GREEN | BLUE |
| RED | BLUE | YELLOW| GREEN |
| RED | BLUE | GREEN | YELLOW|
| RED | GREEN | YELLOW| BLUE |
| RED | GREEN | BLUE | YELLOW|
| YELLOW| RED | BLUE | GREEN |
| YELLOW| RED | GREEN | BLUE |
| YELLOW| BLUE | RED | GREEN |
| YELLOW| BLUE | GREEN | RED |
| YELLOW| GREEN | RED | BLUE |
| YELLOW| GREEN | BLUE | RED |

(You can write these in the blank rows.)

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🔁 Second Question:

What if any car could park in any lot? How many possibilities then?

Now, no restrictions — any car can go in any lot, but still one car per lot.

We are arranging 4 distinct cars into 4 distinct lots → this is just the number of permutations of 4 items:

$$
4! = 4 \times 3 \times 2 \times 1 = 24
$$

✔ So, 24 possibilities if there are no restrictions.

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✔ Final Answers:

- Under original rules (Lot 1 = Red or Yellow only):
→ 12 possible arrangements (we listed all 12 above)

- If any car can park anywhere:
→ 24 possible arrangements

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📝 Summary:

- The restriction reduces the total from 24 to 12.
- This is because only 2 choices for Lot 1 (Red or Yellow), and then 3! = 6 for each case → 2 × 6 = 12.

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Let me know if you'd like this printed in a table format or want it simplified for younger students!