Total number of ways in which six '+' and four '–' sings can be arranged in a line such that no two '–' sings occur together, is

Options

A. 35
B. 18
C. 15
D. 42

Answer

To find the total number of ways to arrange six '+' signs and four '–' signs in a line such that no two '–' signs occur together, we can follow these steps:

### Step 1: Arrange the '+' Signs
First, we arrange the six '+' signs in a row. The arrangement of these signs creates gaps where the '–' signs can be placed.

**Arrangement of '+' signs:**
```
+ + + + + +
```

### Step 2: Identify the Gaps
After arranging the six '+' signs, we can identify the gaps where the '–' signs can be placed. There are gaps before the first '+', between each pair of '+' signs, and after the last '+'.

**Gaps created:**
```
_ + _ + _ + _ + _ + _ +
```
This gives us a total of **7 gaps** (1 before the first '+', 5 between the '+' signs, and 1 after the last '+').

### Step 3: Choose Gaps for '–' Signs
We need to place the four '–' signs in these 7 gaps. The condition is that no two '–' signs can occupy the same gap, ensuring that they do not come together.

### Step 4: Calculate the Combinations
To find the number of ways to choose 4 gaps from the 7 available gaps, we use the combination formula:

\[ \text{Number of ways} = \binom{7}{4} \]

### Step 5: Compute the Combination
The combination formula is given by:

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

For our case:

\[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4! \cdot 3!} \]

Calculating this gives:

\[ = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35 \]

### Final Answer
Thus, the total number of ways to arrange six '+' signs and four '–' signs such that no two '–' signs occur together is **35**.

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