Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all

Q: Wait, why can't we just subtract the case where all 5 girls are together from the total?

That would only remove the case where all 5 are together. It wouldn't account for the specific requirement of 'exactly 4' girls being together. We need to construct the 'exactly 4' scenario from scratch using the Gap Method to ensure the 5th girl is isolated.

Q: Why do we use ${}^6P_2$ instead of ${}^6C_2$ when placing the girl groups?

Great question! The two girl-entities—the block of 4 girls and the single girl—are distinct. Placing the block in gap 1 and the single girl in gap 2 is different from placing the single girl in gap 1 and the block in gap 2. Since the order matters, we use permutations (${}^6P_2$).

Q: Does the internal arrangement of the 5 boys matter?

Absolutely! The boys are distinct individuals. Arranging them in $5!$ ways is the foundation of our Gap Method. Without this, we would be ignoring the permutations of the boys themselves.

Question

Let $n$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then the value of $\frac{m}{n}$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a bustling corridor, tasked with arranging 5 boys and 5 girls in a single, orderly queue. We are looking for two values: $n$, the number of ways all 5 girls stand together, and $m$, the number of ways exactly 4 girls stand together.

Our goal is to find the ratio $\frac{m}{n}$.

### The String Method and the Value of $n$

When we need items to stay together, we use the 'String Method'. Imagine taking a piece of string and tying all 5 girls into one unbreakable unit.

Now, instead of 10 individuals, we have 5 boys and 1 'girl-block', giving us 6 units to arrange. These 6 units can be shuffled in $6!$ ways.

Within their block, the 5 girls can rearrange themselves in $5!$ ways. Thus, the total number of ways for $n$ is:

$$n = 6! 	imes 5!$$

### The Gap Method and the Value of $m$

Now, the challenge shifts. We need exactly 4 girls together, which implies we have two distinct girl-entities: a block of 4 girls and 1 single girl.

To ensure the block of 4 and the single girl are not adjacent, we use the 'Gap Method'. First, we arrange the 5 boys, which can be done in $5!$ ways.

These 5 boys create 6 gaps—one at each end and four between them. We must place our two girl-entities into these 6 gaps.

First, we select 4 girls out of 5 to form the block, which is $\binom{5}{4}$ ways, and arrange them internally in $4!$ ways. Now, we have two distinct items: the block of 4 and the single girl.

We place them into 2 of the 6 available gaps. Since the items are distinct, the order of placement matters, so we use permutations: $P(6, 2)$.

Multiplying these independent choices gives us:

$$m = 5! 	imes \binom{5}{4} 	imes 4! 	imes P(6, 2)$$

### The Grand Synthesis

Let us simplify $m$. We know $\binom{5}{4} = 5$ and $P(6, 2) = 6 	imes 5 = 30$.

Substituting these, we get:

$$m = 5! 	imes (5 	imes 4!) 	imes 30$$

Since $5 	imes 4! = 5!$, the expression simplifies to:

$$m = 5! 	imes 5! 	imes 30$$

Now, we calculate the ratio $\frac{m}{n}$:

$$\frac{m}{n} = \frac{5! 	imes 5! 	imes 30}{6! 	imes 5!}$$

The $5!$ terms cancel out, leaving:

$$\frac{m}{n} = \frac{5! 	imes 30}{6!}$$

Since $6! = 6 	imes 5!$, the expression becomes:

$$\frac{m}{n} = \frac{5! 	imes 30}{6 	imes 5!} = \frac{30}{6} = 5$$

The complexity dissolves into a simple integer. The final result is **5**.

Enter Numerical Value:

Visualized Solution

Understanding the Problem

Calculating $n$ : The String Method

Arranging the Units for $n$

Internal Arrangement for $n$

Understanding $m$ : Exactly 4 Girls Together

The Gap Method for $m$

Identifying the Gaps

Forming the Girl Groups

Placing the Groups in Gaps

Total Ways for $m$

Simplifying $m$

Rearranging Terms in $m$

Final Calculation of $\frac{m}{n}$

The Sigma Insight: Linear Permutations

Analyzing the Setup

The String Method and the Value of $n$

The Gap Method and the Value of $m$

The Grand Synthesis

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Enter Numerical Value:

Visualized Solution

Understanding the Problem

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n: The String Method

Arranging the Units for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Internal Arrangement for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Understanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m: Exactly 4 Girls Together

The Gap Method for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m