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3 years ago

6

How many ways are there to place 3 indistinguishable red chips, 3 indistinguishable blue chips, and 3 indistinguishable green ch

ips in the squares of a 3 x 3 grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
A. 12,
B. 18,
C. 24,
D. 30 ,
E. 36

1 answer:

cricket20 [7]3 years ago

6 0

Answer:

E. 36

Step-by-step explanation:

Let:

$C_1 =$ color 1

$C_2 =$ color 2

$C_3 =$ color 3

Case I

C₂ C₃ C₂

C₃ C₁ C₃

C₁ C₂ C₁

Since C₁ has 3 choices, then C₂ has 2 choices and C₁ has 1 choice

number of combinations = 3 × 2 × 1 = 6

Case II

C₂ C₃ C₁

C₃ C₁ C₂

C₂ C₃ C₁

Since C₁ has 3 choices, then C₂ has 2 choices and C₁ has 1 choice

number of combinations = 3 × 2 × 1 = 6

Case III

C₁ C₃ C₂

C₂ C₁ C₃

C₃ C₂ C₁

Since C₁ has 3 choices, then C₂ has 2 choices and C₁ has 1 choice

number of combinations = 3 × 2 × 1 = 6

Case IV

C₂ C₃ C₁

C₃ C₁ C₂

C₁ C₂ C₃

Since C₁ has 3 choices, then C₂ has 2 choices and C₁ has 1 choice

number of combinations = 3 × 2 × 1 = 6

Case V

C₁ C₂ C₁

C₃ C₁ C₃

C₂ C₃ C₂

Since C₁ has 3 choices, then C₂ has 2 choices and C₁ has 1 choice

number of combinations = 3 × 2 × 1 = 6

Case VI

C₁ C₂ C₃

C₃ C₁ C₂

C₁ C₂ C₃

Since C₁ has 3 choices, then C₂ has 2 choices and C₁ has 1 choice

number of combinations = 3 × 2 × 1 = 6

Therefore, the total combinations = 6 + 6 + 6 + 6 + 6 + 6

= 36

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