There are 2 white, 3 black and 4 red balls. The number of ways in which these balls can be arranged so that no two red balls may occupy consecutive positions is given by……. The balls of some colour are identical.

Answer

To solve the problem of arranging 2 white, 3 black, and 4 red balls such that no two red balls occupy consecutive positions, we can follow these steps:

### Step 1: Arrange the Non-Red Balls
First, we will arrange the 2 white and 3 black balls. Since the balls of the same color are identical, we can use the formula for permutations of multiset:

\[ \text{Number of arrangements} = \frac{n!}{n_1! \cdot n_2!} \]

where \( n \) is the total number of balls, and \( n_1, n_2 \) are the counts of each type of identical ball.

Here, we have:
- Total balls = 2 (white) + 3 (black) = 5
- Identical balls: 2 white and 3 black

Thus, the number of arrangements of the white and black balls is:

\[ \text{Arrangements} = \frac{5!}{2! \cdot 3!} \]

Calculating this gives:

\[ = \frac{120}{2 \cdot 6} = \frac{120}{12} = 10 \]

### Step 2: Determine Positions for Red Balls
Next, we need to place the 4 red balls in such a way that no two red balls are adjacent. After arranging the 5 non-red balls, we can visualize the arrangement as follows:

- The arrangement of 5 non-red balls creates 6 potential slots for the red balls (before the first ball, between the balls, and after the last ball):

```
_ W _ W _ B _ B _ B _
```

Here, "_" represents a potential position for a red ball.

### Step 3: Choose Positions for Red Balls
We need to select 4 out of these 6 slots to place the red balls. The number of ways to choose 4 slots from 6 is given by the combination formula:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

In our case, \( n = 6 \) and \( r = 4 \):

\[ \text{Ways to choose positions} = \binom{6}{4} = \frac{6!}{4! \cdot 2!} = \frac{6 \cdot 5}{2 \cdot 1} = 15 \]

### Step 4: Calculate Total Arrangements
Finally, we multiply the number of arrangements of the non-red balls by the number of ways to place the red balls:

\[ \text{Total arrangements} = \text{Arrangements of non-red balls} \times \text{Ways to choose positions for red balls} \]

\[ = 10 \times 15 = 150 \]

### Final Answer
Thus, the total number of ways to arrange the balls such that no two red balls are adjacent is **150**.

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