The total number of ways in which six '+' and four '-' signs can be arranged in a line such that no two signs '-' occur together, is ……….. .

Explanation

To solve the problem of arranging six '+' signs and four '-' signs such that no two '-' signs occur together, we can follow these steps:

### Step-by-Step Solution:

1. **Arrange the '+' Signs**:
First, we arrange the six '+' signs in a line. The arrangement of these signs creates gaps where the '-' signs can be placed.

- Arrangement: `+ + + + + +`

2. **Identify the Gaps**:
After arranging the '+' signs, we can identify the gaps where the '-' signs can be placed. For six '+' signs, there are seven gaps:
- Gaps: `_ + _ + _ + _ + _ + _ + _`
(The underscores represent the gaps where we can place the '-' signs.)

3. **Select Gaps for '-' Signs**:
We need to choose 4 out of these 7 gaps to place the '-' signs. This ensures that no two '-' signs are adjacent to each other.

- Number of ways to choose 4 gaps from 7:
\[ \binom{7}{4} \]

4. **Calculate the Combinations**:
The combination formula \(\binom{n}{r}\) is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
For our case:
\[ \binom{7}{4} = \frac{7!}{4! \cdot (7-4)!} = \frac{7!}{4! \cdot 3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

5. **Final Answer**:
Since the '-' signs are identical, we do not need to multiply by any factorial for their arrangement. Thus, the total number of ways to arrange the signs is:
\[ \text{Total Ways} = 35 \]

### Final Answer:
The total number of ways in which six '+' and four '-' signs can be arranged in a line such that no two '-' signs occur together is **35**.