Six boys and six girls sit in a row randomly. Find the probability that (i) the six girls sit togeth

Q: Wait, why do we multiply by $6!$ inside the bundle for the first part?

That is a fantastic question! Remember, the girls are distinct individuals. Even if they are tied together in a bundle, they are not identical statues. They can swap seats among themselves in $6!$ ways. We must account for both the arrangement of the 'bundle' relative to the boys and the internal arrangement of the girls within that bundle.

Q: Why do we multiply by 2 in the alternating case?

This is the classic 'trap' of the problem! There are two distinct, mutually exclusive patterns that satisfy the alternating condition: either a boy starts the row (B-G-B-G...) or a girl starts the row (G-B-G-B...). Both are equally valid and must be counted. If you miss the second pattern, you miss half the favorable outcomes!

Q: Can I use combinations instead of permutations to solve this?

You absolutely can! You could choose the positions for the girls first using combinations, like $\binom{12}{6}$, and then arrange the girls in those chosen spots and the boys in the remaining spots. However, for row arrangements, permutations are often more direct and less prone to calculation errors. Both methods will lead you to the same beautiful result.

Question

Six boys and six girls sit in a row randomly. Find the probability that (i) the six girls sit together (ii) the boys and girls sit alternately.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In the world of combinatorics, our universe is defined by the total number of ways twelve distinct individuals—six boys and six girls—can arrange themselves in a line. Before we impose any constraints, the total number of arrangements is given by:

$$N = 12!$$

This value serves as the denominator for our probability calculations, representing the entire sample space of possibilities.

### The Art of the Bundle: Solving Part (i)

To satisfy the constraint that all six girls must sit together, we employ the **String Method**. We treat the six girls as a single "super-unit" or bundle.

Now, we arrange the six boys and the one girl-bundle, which gives us $6 + 1 = 7$ units. These units can be arranged in $7!$ ways. Within the bundle, the six girls can rearrange themselves in $6!$ ways.

By the Fundamental Principle of Counting, the number of favorable arrangements is:

$$n(E_1) = 7! 	imes 6!$$

The probability is the ratio of favorable outcomes to the total sample space:

$$P(	ext{Girls together}) = \frac{7! 	imes 6!}{12!}$$

Expanding $12!$ as $12 	imes 11 	imes 10 	imes 9 	imes 8 	imes 7!$, the $7!$ terms cancel out. This simplifies to:

$$P(	ext{Girls together}) = \frac{6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1}{12 	imes 11 	imes 10 	imes 9 	imes 8} = \mathbf{\frac{1}{132}}$$

### The Symmetry of Alternation: Solving Part (ii)

For the alternating arrangement, we must consider two mutually exclusive scenarios (universes) where boys and girls occupy alternating seats.

In **Universe A**, the pattern is $B-G-B-G-B-G-B-G-B-G-B-G$. The boys occupy the six odd positions and the girls occupy the six even positions, resulting in $6! 	imes 6!$ arrangements.

In **Universe B**, the pattern is $G-B-G-B-G-B-G-B-G-B-G-B$. This also yields $6! 	imes 6!$ arrangements. Since these scenarios are distinct, we sum them:

$$n(E_2) = 2 	imes (6! 	imes 6!)$$

The probability is calculated as follows:

$$P(	ext{Alternate}) = \frac{2 	imes 6! 	imes 6!}{12!}$$

Expanding $12!$ to $12 	imes 11 	imes 10 	imes 9 	imes 8 	imes 7 	imes 6!$ allows one $6!$ to cancel. After performing the arithmetic, we arrive at the final result:

$$P(	ext{Alternate}) = \mathbf{\frac{1}{462}}$$

### The Mentor's Reflection

Probability is not merely about memorizing formulas; it is about visualizing constraints. When you encounter "together," think of bundles; when you encounter "alternately," think of patterns and symmetry.

You have successfully navigated the sample space, mastered the bundling technique, and respected the symmetry of alternating patterns. Continue to seek the underlying structure in every problem you face.

Animated Solution for Mathematics - Probability: Six boys and six girls sit in a row randomly. Find the probability that (i) the six girls sit together (ii) the boys and girls sit alternately.

Visualized Solution

Defining the Sample Space

Part (i): The String Method

Arranging the Bundle

Calculating Probability for Part (i)

Part (ii): Alternating Arrangement (Pattern 1)

Alternating Arrangement (Pattern 2)

Calculating Probability for Part (ii)

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Art of the Bundle

The Symmetry of Alternation

The Mentor's Reflection

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