Understanding the Row Arrangement Problem

The question asks about the probability of a specific arrangement when six people are placed in a row for a group photograph. We have three boys (P, Q, R) and three girls (S, T, U), making a total of six individuals. We need to find the probability that two specific boys, P and Q, occupy the two end positions in the row.

Calculating Total Possible Arrangements

First, let's determine the total number of ways to arrange the six individuals in a row. When arranging 'n' distinct items in a row, the total number of arrangements is given by n! (n factorial).

• Total number of people = 6
• Total possible arrangements = \(6!\)
• \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\)

So, there are 720 different ways to arrange the six people in a row.

Calculating Favorable Arrangements (P and Q at Ends)

Next, we need to find the number of arrangements where P and Q are at the two end positions. The two end positions can be occupied by P and Q in two ways:

• Case 1: P is at the left end and Q is at the right end.
• Case 2: Q is at the left end and P is at the right end.

Let's consider Case 1: P at the left end, Q at the right end.

_ P _ _ _ Q _

The remaining four positions in the middle must be filled by the other four individuals (R, S, T, U). The number of ways to arrange these four individuals in the four middle positions is \(4!\).

• Number of ways for Case 1 = \(4! = 4 \times 3 \times 2 \times 1 = 24\)

Now let's consider Case 2: Q at the left end, P at the right end.

_ Q _ _ _ P _

Similarly, the remaining four positions must be filled by the other four individuals (R, S, T, U). The number of ways to arrange these four individuals in the four middle positions is \(4!\).

• Number of ways for Case 2 = \(4! = 4 \times 3 \times 2 \times 1 = 24\)

The total number of favorable arrangements (where P and Q are at the ends) is the sum of the ways for Case 1 and Case 2.

• Total favorable arrangements = Number of ways for Case 1 + Number of ways for Case 2
• Total favorable arrangements = \(24 + 24 = 48\)

Calculating the Probability

The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.

Probability (P and Q at ends) = \(\frac{\text{Number of favorable arrangements}}{\text{Total number of possible arrangements}}\)

Probability = \(\frac{48}{720}\)

Now, we simplify the fraction:

• \(\frac{48}{720} = \frac{24 \times 2}{24 \times 30} = \frac{2}{30} = \frac{1}{15}\)

Thus, the probability that P and Q take the two end positions is \(\frac{1}{15}\).

Revision Table: Key Concepts in Probability

Additional Information on Arrangements and Probability

This problem is a classic example of calculating probability using permutations (arrangements). The key steps involved are identifying the total possible outcomes and the number of specific outcomes that satisfy the condition (favorable outcomes).

• Total Outcomes: This is the total number of ways the entire group can be arranged without any restrictions. Since there are 6 distinct individuals, the total arrangements are \(6!\).
• Favorable Outcomes: Here, the restriction is that P and Q must be at the ends. We fix P and Q at the two end spots and then arrange the remaining individuals in the middle. Since P and Q can swap positions at the ends (P-Q or Q-P), we consider these two possibilities. The remaining 4 people can be arranged in the 4 middle spots in \(4!\) ways. So, the total favorable arrangements are \(2 \times 4!\).
• Calculating Probability: Once both total and favorable outcomes are found, the probability is simply the ratio. Simplifying the resulting fraction is important to get the final answer in its simplest form.

Understanding the difference between permutations (where order matters) and combinations (where order does not matter) is crucial in solving such problems. In this question, the arrangement in a row implies that the order of individuals is important, hence we use permutations (factorial calculations).