You are given `8` balls of different colour (black,white ......). The number of ways in which these balls can be arranged in a row so that the two balls of particular colour (say red and white) may never come together is-

Options

A. 8!-2.7!
B. 6.7!
C. 2.6!`""^(7)C_(2)`
D. none

Answer

To solve the problem of arranging 8 balls of different colors such that two specific balls (red and white) do not come together, we can follow these steps:

### Step 1: Calculate the total arrangements without restrictions
First, we calculate the total number of arrangements of the 8 balls without any restrictions. Since all the balls are of different colors, the total arrangements can be calculated using the factorial of the number of balls.

\[ \text{Total arrangements} = 8! = 40320 \]

**Hint:** Remember that the factorial of a number \( n \) (denoted \( n! \)) is the product of all positive integers up to \( n \).

### Step 2: Calculate arrangements where red and white are together
Next, we treat the red and white balls as a single unit or block. This means that instead of 8 separate balls, we now have 7 units to arrange (the block of red and white, plus the other 6 balls).

The arrangements of these 7 units can be calculated as follows:

\[ \text{Arrangements of 7 units} = 7! = 5040 \]

Since the red and white balls can be arranged within their block in 2 different ways (red first or white first), we multiply the arrangements of the 7 units by the arrangements of the red and white balls.

\[ \text{Arrangements with red and white together} = 7! \times 2! = 5040 \times 2 = 10080 \]

**Hint:** When treating two items as a single unit, remember to multiply by the arrangements of those two items.

### Step 3: Calculate arrangements where red and white are not together
To find the arrangements where the red and white balls are not together, we subtract the arrangements where they are together from the total arrangements.

\[ \text{Arrangements where red and white are not together} = 8! - (7! \times 2!) \]

Substituting the values we calculated:

\[ \text{Arrangements where red and white are not together} = 40320 - 10080 = 30240 \]

**Hint:** The principle of complementary counting can be very useful: subtract the unwanted arrangements from the total arrangements.

### Final Answer
Thus, the number of ways to arrange the 8 balls such that the red and white balls do not come together is:

\[ \boxed{30240} \]