Probability of Specific Individuals Sitting Together

The question asks for the probability that two specific individuals, Q (a boy) and U (a girl), sit together when six people (three boys P, Q, R and three girls S, T, U) are arranged in a row for a group photograph.

Calculating Total Possible Arrangements

We have a total of 6 people to arrange in a row. The number of ways to arrange 6 distinct people in a row is given by the factorial of 6.

Total number of arrangements = \(6!\)

\(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\)

So, there are 720 total possible arrangements for the group photograph.

Calculating Favorable Arrangements (Q and U Sit Together)

We want to find the number of arrangements where Q and U sit next to each other. To do this, we can treat Q and U as a single unit or a block. Now, instead of arranging 6 individual people, we are arranging 5 units:

• The block containing Q and U (QU)
• The remaining 4 people: P, R, S, T

These 5 units can be arranged in \(5!\) ways.

Number of ways to arrange the 5 units = \(5!\)

\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

However, within the block containing Q and U, the two people can arrange themselves in two ways: Q followed by U (QU) or U followed by Q (UQ). The number of ways Q and U can arrange themselves within their block is \(2!\).

Number of ways Q and U can arrange within the block = \(2!\)

\(2! = 2 \times 1 = 2\)

To find the total number of arrangements where Q and U sit together, we multiply the number of ways to arrange the 5 units by the number of ways Q and U can arrange themselves within their block.

Total number of favorable arrangements = (Number of ways to arrange the 5 units) \(\times\) (Number of ways Q and U arrange within the block)

Total number of favorable arrangements = \(5! \times 2!\)

Total number of favorable arrangements = \(120 \times 2 = 240\)

There are 240 arrangements where Q and U sit together.

Calculating the Probability

The probability that Q and U sit together is the ratio of the number of favorable arrangements to the total number of possible arrangements.

Probability (Q and U sit together) = \(\frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements}}\)

Probability = \(\frac{240}{720}\)

We can simplify this fraction:

\(\frac{240}{720} = \frac{24}{72} = \frac{1}{3}\)

Alternatively, using factorials directly:

Probability = \(\frac{5! \times 2!}{6!}\)

Probability = \(\frac{(5 \times 4 \times 3 \times 2 \times 1) \times (2 \times 1)}{(6 \times 5 \times 4 \times 3 \times 2 \times 1)}\)

Probability = \(\frac{5! \times 2}{6 \times 5!}\)

We can cancel out \(5!\) from the numerator and the denominator:

Probability = \(\frac{2}{6}\)

Probability = \(\frac{1}{3}\)

The probability that Q and U sit together is \(\frac{1}{3}\).

Conclusion

The probability that Q and U sit together is \(\frac{1}{3}\).

Revision Table: Probability and Arrangements

Additional Information: Types of Arrangement Problems

Understanding different types of arrangement problems is key in probability and combinatorics.

• Linear Arrangements: Arranging items in a straight line, as in this problem. The total number of arrangements of \(n\) distinct items is \(n!\).
• Circular Arrangements: Arranging items in a circle. The number of arrangements of \(n\) distinct items in a circle is \((n-1)!\) if rotations are considered the same.
• Arrangements with Repetition: If items are repeated, the formula changes. For example, arranging letters in the word "MISSISSIPPI".
• Arrangements with Constraints: Problems like the one solved here, where specific items must sit together or apart, or specific positions are required.

In this problem, the constraint is that Q and U must sit together, which is handled by the "treating as a unit" method.