There are four pairs of shoes of different sizes. Each of the 8 shoes can be coloured with one of the four colours. Black, Brown, White and Red. The number of ways the shoes can be coloured so that in atleast three pairs, the left shoe and the right shoe do not have the same colour is

Options

A. `12^(4)`
B. `28xx12^(3)`
C. `16xx12^(3)`
D. `4xx12^(3)`

Answer

To solve the problem of coloring the shoes such that at least 3 pairs have different colors for the left and right shoes, we can break down the solution step by step.

### Step 1: Understanding the Problem
We have 4 pairs of shoes, which means there are 8 individual shoes. Each shoe can be colored in one of 4 colors: Black, Brown, White, and Red. We need to find the number of ways to color the shoes such that at least 3 pairs have different colors for the left and right shoes.

### Step 2: Total Ways to Color the Shoes
First, let's calculate the total number of ways to color all 8 shoes without any restrictions. Each shoe can be one of 4 colors, so the total number of ways to color all shoes is:
\[ 4^8 \]

### Step 3: Counting the Cases
Now, we need to consider the cases where at least 3 pairs have different colors. We can break this down into two cases:
1. Exactly 3 pairs have different colors.
2. All 4 pairs have different colors.

### Step 4: Case 1 - Exactly 3 Pairs with Different Colors
To find the number of ways to color the shoes such that exactly 3 pairs have different colors, we can:
1. Choose 3 pairs from the 4 pairs. The number of ways to choose 3 pairs from 4 is given by:
\[ \binom{4}{3} = 4 \]
2. For each of the 3 chosen pairs, we can color them in different colors. The first shoe can be colored in any of the 4 colors, and the second shoe must be a different color (3 choices). Thus, for each of the 3 pairs:
\[ 4 \times 3 = 12 \text{ ways} \]
Therefore, for 3 pairs, the total ways is:
\[ 12^3 \]
3. The remaining pair can be colored in any of the 4 colors (both shoes can be the same color). Thus, there are:
\[ 4 \text{ ways} \]

Combining these, the total number of ways for this case is:
\[ \text{Total for Case 1} = \binom{4}{3} \times 12^3 \times 4 = 4 \times 12^3 \times 4 = 16 \times 12^3 \]

### Step 5: Case 2 - All 4 Pairs with Different Colors
For the case where all 4 pairs have different colors:
1. All pairs can be colored in the same way as in the previous case, but now we have 4 pairs. The first shoe can be colored in any of the 4 colors, and the second shoe must be a different color (3 choices). Thus, for each of the 4 pairs:
\[ 4 \times 3 = 12 \text{ ways} \]
Therefore, for 4 pairs, the total ways is:
\[ 12^4 \]

### Step 6: Combining Both Cases
Now, we combine the results from both cases to find the total number of ways to color the shoes such that at least 3 pairs have different colors:
\[ \text{Total} = 16 \times 12^3 + 12^4 \]

### Step 7: Final Calculation
To simplify:
\[ 12^4 = 12 \times 12^3 \]
Thus:
\[ \text{Total} = 16 \times 12^3 + 12 \times 12^3 = (16 + 12) \times 12^3 = 28 \times 12^3 \]

### Conclusion
The number of ways to color the shoes such that at least 3 pairs have different colors is:
\[ \boxed{28 \times 12^3} \]