Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the w

Q: Wait, why do we use permutations ($^nP_r$) instead of combinations ($^nC_r$)?

This is a fantastic question! The key lies in the phrase 'numbered chairs'. If the chairs were identical, swapping two people would not create a new arrangement. But because the chairs are numbered $1$ through $8$, chair $1$ is fundamentally different from chair $2$. Therefore, seating Person A in chair $1$ and Person B in chair $2$ is a distinct arrangement from seating Person B in chair $1$ and Person A in chair $2$. Whenever the 'where' matters, we use permutations.

Q: Why are there $6$ chairs left for the men? Shouldn't it be $4$ chairs (chairs $5$ to $8$)?

I love that you are thinking about the constraints! It is a common trap to think the men are restricted to chairs $5$ through $8$. However, the problem only says the women must choose from chairs $1$ to $4$. Once the women are seated, the men can sit in *any* of the remaining chairs. Since there were $8$ chairs total and $2$ are now occupied, there are $8 - 2 = 6$ chairs available for the men. This includes the $2$ empty chairs from the first zone and all $4$ chairs from the second zone.

Q: Does the order of seating (women first, then men) actually change the final answer?

In this specific problem, the constraint forces us to seat the women first. If we were to seat the men first, the calculation would become much more complex because the number of available chairs for the women would depend on how many men chose chairs from the $1$ to $4$ range. By following the problem's sequential structure, we simplify the logic significantly. Always tackle the most restricted group first!

Question

Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is

Options:

(A) $^6C_3 	imes ^4C_2$

(B) $^4P_2 	imes ^6P_3$

(C) $^4C_2 + ^4P_3$

(D) none of these

Sigma Percentile · Accepted Answer

### The Geometry of Choice: A Combinatorial Journey

Welcome, fellow traveler, to the elegant world of combinatorics. Today, we are not just solving a problem; we are mapping out possibilities.

We have eight chairs, numbered $1$ through $8$, and a group of five people—two women and three men—who are eager to take their seats. This might seem like a simple seating arrangement, but it is a masterclass in understanding constraints and the Fundamental Principle of Counting.

### Phase 1: The Power of Distinct Entities

First, let us ground ourselves. Why do we care that the chairs are numbered? In the realm of counting, this is the difference between a combination and a permutation.

If the chairs were identical, we would only care about which chairs were chosen. But here, chair $1$ is a unique location, distinct from chair $2$.

Because the chairs are labeled, the order in which our friends sit matters. This immediately tells us that we are dealing with permutations. We are not just selecting seats; we are assigning people to specific, unique positions.

### Phase 2: The Women's Constraint

Every complex problem has a 'bottleneck'—a constraint that dictates the flow of the solution. Here, the women have a specific requirement: they must choose their chairs from the subset of chairs marked $1$ through $4$. Let us call this the 'Blue Zone'.

Imagine you are the women. You have $4$ chairs to choose from, and you need to occupy $2$ of them. Since the chairs are distinct, the number of ways to arrange $2$ women in $4$ chairs is given by the permutation formula:

$$^nP_r = \frac{n!}{(n-r)!}$$

Substituting our values, we get:

$$^4P_2 = \frac{4!}{(4-2)!} = \frac{4 	imes 3 	imes 2!}{2!} = 12$$

There are exactly $12$ ways for the women to claim their spots in the Blue Zone.

### Phase 3: The Men's Turn and the Remaining Pool

Now, here is where many students stumble. We have seated the women, and they have occupied $2$ chairs. We started with $8$ chairs. How many are left?

It is a simple subtraction: $8 - 2 = 6$. These $6$ chairs are the entire remaining pool. It does not matter if they were in the 'Blue Zone' or the 'Red Zone' (chairs $5$ to $8$); they are all just empty chairs now.

We have $3$ men who need to be seated in these $6$ available chairs. Again, because the chairs are distinct, we use permutations:

$$^6P_3 = \frac{6!}{(6-3)!} = \frac{6!}{3!} = 6 	imes 5 	imes 4 = 120$$

There are $120$ ways for the men to take their seats.

### Phase 4: The Synthesis

We are almost there. We have calculated the ways for the women ($12$) and the ways for the men ($120$). To find the total number of arrangements, we invoke the Fundamental Principle of Counting.

Since these are two sequential, independent events that together complete the task, we multiply the possibilities:

$$12 	imes 120 = 1440$$

Mathematically, this is expressed as $^4P_2 	imes ^6P_3$. It is a beautiful, clean result.

We handled the restriction first, cleared the path, and then allowed the rest of the group to fill the remaining space. Remember this strategy: whenever you face a problem with constraints, identify the restricted group, solve for them first, and then watch the rest of the problem fall into place.

The final answer is **1440**.

Select Answer:

Visualized Solution

Visualizing the Setup

The Women's Constraint

Seating the Women: $^{4} P_{2}$

Calculating $^{4} P_{2}$

Identifying Remaining Chairs

Seating the Men: $^{6} P_{3}$

Calculating $^{6} P_{3}$

Total Arrangements

Final Answer

The Sigma Insight: Linear Permutations

The Geometry of Choice

Phase 1

Phase 2

Phase 3

Phase 4

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