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There are four pairs of shoes of differe...

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

A

`12^(4)`

B

`28xx12^(3)`

C

`16xx12^(3)`

D

`4xx12^(3)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the number of ways to color 8 shoes (4 pairs) such that in at least 3 pairs, the left and right shoes do not have the same color. The colors available are Black, Brown, White, and Red. Let's break down the solution step by step: ### Step 1: Total Coloring Options Each shoe can be colored in one of 4 colors. Therefore, the total number of ways to color all 8 shoes without any restrictions is: \[ 4^8 \] ### Step 2: Cases to Consider We need to consider two cases: 1. Exactly 3 pairs have different colors for the left and right shoes. 2. All 4 pairs have different colors for the left and right shoes. ### Step 3: Case 1 - Exactly 3 Pairs with Different Colors - **Choosing 3 pairs**: We can choose 3 pairs out of 4 in \( \binom{4}{3} = 4 \) ways. - **Coloring the chosen pairs**: For each of the 3 pairs, the left and right shoes can be colored in different colors. The left shoe can be any of the 4 colors, and the right shoe can be any of the remaining 3 colors. Therefore, for 3 pairs, the number of ways to color them is: \[ 4 \times 3 \text{ (for the first pair)} \times 4 \times 3 \text{ (for the second pair)} \times 4 \times 3 \text{ (for the third pair)} = (4 \times 3)^3 = 12^3 \] - **Coloring the remaining pair**: The last pair can be colored in any of the 4 colors for both shoes, so there are \( 4^2 = 16 \) ways to color the remaining pair. Thus, the total number of ways for this case is: \[ \binom{4}{3} \times 12^3 \times 16 = 4 \times 12^3 \times 16 \] ### Step 4: Case 2 - All 4 Pairs with Different Colors - **Choosing all pairs**: Here, all pairs will have different colors. We can color the left shoe of each pair in any of the 4 colors, and the right shoe can be any of the remaining 3 colors. Thus, for each of the 4 pairs: \[ (4 \times 3)^4 = 12^4 \] ### Step 5: Total Ways for All Cases Now, we sum the number of ways from both cases: \[ \text{Total Ways} = (4 \times 12^3 \times 16) + 12^4 \] ### Step 6: Simplifying the Expression We can factor out \( 12^3 \): \[ \text{Total Ways} = 12^3 (4 \times 16 + 12) = 12^3 (64 + 12) = 12^3 \times 76 \] ### Final Calculation Calculating \( 12^3 \): \[ 12^3 = 1728 \] Thus, the total number of ways is: \[ 1728 \times 76 = 131328 \] ### Conclusion The number of ways to color the shoes such that in at least 3 pairs, the left shoe and the right shoe do not have the same color is: \[ \boxed{131328} \]
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