In how many ways can 7 red and 6 black balls be arranged in a row, if no two black balls are together?

Answer

To solve the problem of arranging 7 red and 6 black balls in a row such that no two black balls are together, we can follow these steps:

### Step 1: Arrange the Red Balls
First, we need to arrange the 7 red balls. When we arrange these red balls in a row, they will create gaps where the black balls can be placed.

**Arrangement of Red Balls:**
- R R R R R R R (7 red balls)

### Step 2: Identify the Gaps
Once the red balls are arranged, we can identify the gaps where the black balls can be placed. There will be gaps before the first red ball, between the red balls, and after the last red ball.

**Gaps created:**
- _ R _ R _ R _ R _ R _ R _ R _
- This gives us a total of 8 gaps (1 before the first red ball, 6 between the red balls, and 1 after the last red ball).

### Step 3: Choose Gaps for Black Balls
We need to place 6 black balls in these 8 gaps, ensuring that no two black balls are placed in the same gap (to satisfy the condition that no two black balls are together).

**Choosing Gaps:**
- We need to choose 6 gaps out of the 8 available gaps to place the black balls. The number of ways to choose 6 gaps from 8 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of gaps and \( r \) is the number of gaps to choose.

### Step 4: Calculate the Combinations
The number of ways to choose 6 gaps from 8 is calculated as follows:

\[ \binom{8}{6} = \frac{8!}{6! \cdot (8-6)!} = \frac{8!}{6! \cdot 2!} \]

### Step 5: Simplify the Expression
Now, we can simplify the expression:

\[ \binom{8}{6} = \frac{8 \times 7}{2 \times 1} = \frac{56}{2} = 28 \]

### Final Answer
Thus, the total number of ways to arrange 7 red and 6 black balls in a row such that no two black balls are together is **28**.

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