Total number of ways in which six '+' and four '-' signs can be arranged in a line such that no two

Q: Wait, why do we include the gaps at the ends? Shouldn't we only count the gaps between the plus signs?

That is a classic trap! If you only count the gaps *between* the plus signs, you are essentially saying that a minus sign cannot be at the very beginning or the very end of the row. But the problem only says no two minus signs can be adjacent. A minus sign at the start or end is perfectly fine, as long as it isn't next to another minus sign. Including the end gaps is essential to cover all valid arrangements.

Q: Why don't we arrange the minus signs? Shouldn't we multiply by $4!$?

Great question! We only arrange items when they are distinct. Here, all the minus signs are identical. If you swap one minus sign with another, the arrangement looks exactly the same. Therefore, there is only $1$ way to arrange the minus signs once you have chosen their positions. We only need to focus on the selection of the gaps.

Q: Can we solve this using the 'Total minus Unwanted' method?

You could, but it would be significantly more complex. You would have to calculate the total arrangements and then subtract cases where at least two minus signs are together, then add back cases where three are together, and so on (using the Principle of Inclusion-Exclusion). The Gap Method is much more elegant and direct for this specific constraint.

Question

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

Sigma Percentile · Accepted Answer

### The Art of Combinatorics: Mastering the Gap Method

Imagine you are tasked with arranging six $+$ signs and four $-$ signs in a line such that no two $-$ signs are adjacent. When you encounter a constraint that forbids adjacency, your mathematical intuition should immediately pivot to the **Gap Method**.

This is a logical strategy that transforms a complex counting problem into a simple act of selection.

### The Foundation: Placing the Unrestricted

We begin by placing our "well-behaved" items—the six $+$ signs. Since these signs are identical, there is only $1$ way to arrange them in a row:
$+ + + + + +$

These six pillars create spaces around them where we can safely place our $-$ signs. Specifically, there is a gap before the first plus, a gap between every pair of pluses, and a gap after the last plus.

For $n$ items, there are $n+1$ available gaps. With $6$ plus signs, we have $6+1=7$ distinct gaps. These gaps serve as the only safe havens for our $-$ signs, ensuring that no two $-$ signs can ever touch.

### The Selection: Choosing the Safe Havens

We have $4$ identical $-$ signs to place into $7$ available gaps. Since the signs are identical, the order of placement does not matter; we only care about which gaps we choose to occupy.

This is a classic combination problem. We must select $4$ gaps out of the $7$ available, which is represented by the binomial coefficient:
$${}^{7}C_{4}$$

This calculation represents the heart of the solution. We are effectively choosing the "homes" for our minus signs within the structure created by the plus signs.

### The Final Calculation: Elegance in Arithmetic

To find the total number of arrangements, we compute ${}^{7}C_{4}$. Using the symmetry property of binomial coefficients, ${}^{n}C_{r} = {}^{n}C_{n-r}$, we know that:
$${}^{7}C_{4} = {}^{7}C_{3}$$

Expanding this expression, we get:
$${}^{7}C_{3} = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1}$$

The denominator $3 	imes 2 	imes 1$ equals $6$. We can neatly cancel this $6$ with the $6$ in the numerator, leaving us with $7 	imes 5$.

The final result is **$35$**. By visualizing the gaps and trusting the logic of combinations, we have successfully navigated the constraints to find the total number of valid arrangements.

Animated Solution for Mathematics - Permutations and Combinations: Total number of ways in which six '+' and four '-' signs can be arranged in a line such that no two '-' signs occur together is .........

Enter Numerical Value:

Visualized Solution

The Gap Method Strategy

Arranging the $+$ Signs

Creating the Gaps

Counting Available Gaps

Selecting Gaps for $-$ Signs

Arranging Identical $-$ Signs

Applying Combinations

Simplifying $^{7} C_{4}$

Calculating $^{7} C_{3}$

Final Computation

The Sigma Insight: Linear Permutations

The Art of Combinatorics

The Foundation

The Selection

The Final Calculation

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