In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, 3

Q: Wait, why do we use permutations ($^{n}P_{r}$) instead of combinations ($^{n}C_{r}$)?

That is a brilliant question! We use permutations because the seats are numbered. In a van, sitting in the front-left seat is physically different from sitting in the back-right seat. Because the seats are distinct, the order in which we place the people matters. If the seats were identical, we would use combinations, but here, every seat has an identity.

Q: Why is the denominator $7 \times 13!$ and not $14!$?

Remember, we are not filling every single seat. We have $14$ seats but only $12$ people. The total number of ways to arrange $12$ people in $14$ seats is $^{14}P_{12}$, which is $\frac{14!}{(14-12)!} = \frac{14!}{2!}$. When you expand $14!$ as $14 \times 13!$, and divide by $2$, you get $7 \times 13!$. It is just a simplified way to write the total sample space.

Q: Why do we multiply by $4$ when choosing the girls' seats?

Let's look at one back row of $4$ seats. The adjacent triplets are $\{1,2,3\}$ and $\{2,3,4\}$. That is $2$ ways per van. Since we have $2$ vans, we have $2 \times 2 = 4$ total ways to choose the block of seats for the girls. It is all about counting the physical possibilities!

Question

In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, 3 in the front and 4 at the back? How many seating arrangements are possible if 3 girls should sit together in a back row on adjacent seats? Now, if all the seating arrangements are equally likely, what is the probability of 3 girls sitting together in a back row on adjacent seats?

Sigma Percentile · Accepted Answer

### The Geometry of the Road Trip: A Combinatorial Journey

Imagine you are standing in a parking lot with two vans. Each van is a masterpiece of engineering, designed with a specific seating layout: $3$ seats in the front and $4$ in the back.

That gives us $7$ seats per van, and with two vans, we have a total of $14$ numbered seats. We have $12$ people—$3$ girls and $9$ boys—waiting to embark on this journey.

The question isn't just about who sits where; it is about the sheer number of ways we can organize this chaos. Welcome to the world of permutations, where every seat has a name and every arrangement tells a story.

### Phase 1: The Total Sample Space

Before we impose any rules, let's look at the raw potential of this scenario. We have $14$ distinct seats and $12$ distinct people.

Because the seats are numbered, the order matters immensely. If Alice sits in seat $1$ and Bob in seat $2$, that is a different arrangement than if Bob sits in seat $1$ and Alice in seat $2$.

This is the classic definition of a permutation. We are choosing $12$ seats out of $14$ and arranging $12$ people in them. Mathematically, this is $^{14}P_{12}$.

Using our formula, $^{n}P_{r} = \frac{n!}{(n-r)!}$, we get:

$$^{14}P_{12} = \frac{14!}{(14-12)!} = \frac{14!}{2!}$$

To make our lives easier later, let's write this as $7 	imes 13!$. Keep this number in your pocket; it is the bedrock of our probability calculation.

### Phase 2: The Constraint - The Girls' Block

Now, let's introduce the constraint. The three girls must sit together in a back row on adjacent seats.

This is where we use the 'Block Method.' We don't see them as three individuals anymore; we see them as a single, inseparable unit that needs to occupy a specific type of space.

Look at the back row of one van. It has $4$ seats. How can we fit a block of $3$ girls?

They can take seats $1, 2, 3$ or seats $2, 3, 4$. That is $2$ possible blocks per van. Since we have $2$ vans, we have $2 	imes 2 = 4$ possible blocks where the girls can sit.

### Phase 3: Arranging the Girls and the Boys

Once we have chosen one of these $4$ blocks, the girls are not just sitting there; they are individuals. They can arrange themselves within those $3$ seats in $3!$ ways, which is $6$ ways.

Now, what about the boys? We have $9$ boys left and $11$ seats remaining ($14$ total seats minus the $3$ occupied by the girls).

We need to arrange these $9$ boys in the $11$ remaining seats. This is $^{11}P_{9}$, which is:

$$^{11}P_{9} = \frac{11!}{(11-9)!} = \frac{11!}{2!}$$

### Phase 4: The Grand Synthesis

Let's multiply these independent choices to find the total favorable arrangements:

$$4 	imes 3! 	imes \frac{11!}{2!} = 4 	imes 6 	imes \frac{11!}{2} = 24 	imes \frac{11!}{2} = 12 	imes 11!$$

And here is the beauty of mathematics: $12 	imes 11!$ is exactly $12!$. The complexity collapses into elegance.

Finally, the probability $P$ is the ratio of favorable outcomes to total outcomes:

$$P = \frac{12!}{7 	imes 13!}$$

Expanding $13!$ as $13 	imes 12!$, we get:

$$P = \frac{12!}{7 	imes 13 	imes 12!} = \frac{1}{7 	imes 13} = \frac{1}{91}$$

There you have it. The chaos of the road trip is tamed by the logic of combinatorics. Remember, whenever you face a constraint, treat the constrained items as a block, arrange them, and then handle the rest. You have got this! The final answer is **$1/91$**.

Visualized Solution

Visualizing the Seating Layout

Calculating Total Arrangements $^{14} P_{12}$

Simplifying the Total Ways

Constraining the Girls' Seats

Arranging the Girls ( $3!$ )

Seating the Nine Boys ( $^{11} P_{9}$ )

Total Favorable Arrangements ( $12!$ )

Calculating the Final Probability $P$

Final Result $P = \frac{1}{91}$

The Sigma Insight: Classical Definition of Probability

The Geometry of the Road Trip

Phase 1

Phase 2

Phase 3

Phase 4

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