Consider the following for the next items that follow: Three boys P, Q, R and three girls S, T, U are to be arranged in a row for a group photograph.
What is the probability that all three boys sit together?
The question asks for the probability that three boys sit together when six people (three boys and three girls) are arranged in a row for a group photograph. To find the probability, we need two values:
Probability is then calculated as:
$\text{Probability} = \frac{\text{Number of favorable arrangements}}{\text{Total number of possible arrangements}}$
We have a total of 6 distinct people (3 boys and 3 girls) to arrange in a row. The number of ways to arrange $n$ distinct items in a row is given by $n!$ (n factorial).
In this case, $n=6$.
Total number of arrangements $= 6!$
$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$
So, there are 720 total possible ways to arrange the six people in a row.
We want to find the number of arrangements where all three boys (P, Q, R) sit together. To handle this condition, we can treat the group of three boys as a single unit.
Imagine tying the three boys together. Now, instead of 6 individual people, we are arranging 4 entities:
The number of ways to arrange these 4 entities in a row is $4!$.
$4! = 4 \times 3 \times 2 \times 1 = 24$
However, within the unit of three boys, the boys themselves can be arranged in different orders. The three boys (P, Q, R) can arrange themselves within their unit in $3!$ ways.
$3! = 3 \times 2 \times 1 = 6$
To find the total number of arrangements where the three boys sit together, we multiply the number of ways to arrange the 4 entities by the number of ways the boys can arrange themselves within their unit.
Number of favorable arrangements = (Arrangement of the 4 entities) $\times$ (Arrangement of boys within the unit)
Number of favorable arrangements $= 4! \times 3! = 24 \times 6 = 144$
So, there are 144 arrangements where all three boys sit together.
Now we can calculate the probability that all three boys sit together using the formula:
$\text{Probability} = \frac{\text{Number of favorable arrangements}}{\text{Total number of possible arrangements}}$
$\text{Probability} = \frac{144}{720}$
We can simplify this fraction:
$\frac{144}{720} = \frac{144 \div 144}{720 \div 144} = \frac{1}{5}$
Alternatively, we can simplify step-by-step:
$\frac{144}{720} = \frac{12 \times 12}{60 \times 12} = \frac{12}{60} = \frac{12 \div 12}{60 \div 12} = \frac{1}{5}$
The probability that all three boys sit together is $\frac{1}{5}$.
| Description | Calculation | Result |
|---|---|---|
| Total number of people | 3 boys + 3 girls | 6 |
| Total possible arrangements of 6 people | $6!$ | 720 |
| Entities to arrange (Boy unit + 3 girls) | 4 | |
| Arrangements of the 4 entities | $4!$ | 24 |
| Arrangements of boys within the unit | $3!$ | 6 |
| Favorable arrangements (boys together) | $4! \times 3!$ | 144 |
| Probability (Favorable / Total) | $\frac{144}{720}$ | $\frac{1}{5}$ |
Based on our calculations, the probability that all three boys sit together is $\frac{1}{5}$. This corresponds to Option 1.
| Concept | Definition/Explanation | Formula/Notation |
|---|---|---|
| Probability | The likelihood of an event occurring. | $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$ |
| Permutation (Arrangement) | An arrangement of items in a specific order. | For arranging $n$ distinct items: $n!$ |
| Factorial | The product of all positive integers up to a given integer $n$. | $n! = n \times (n-1) \times \dots \times 2 \times 1$ |
| Treating items as a unit | A technique used in permutations where a group of items must stay together. Treat the group as one item, arrange, then multiply by the internal arrangements of the group. |
Understanding how to calculate permutations and combinations is crucial for solving probability problems involving arrangements. Here are some related concepts and problem types:
These concepts build upon the fundamental principles of counting and probability, helping solve a wider range of arrangement problems.
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What is the probability that boys and girls sit alternatively?