Circular Permutation Probability Calculation
We need to find the probability that 2 specific students out of 9 always stand together when arranged in a circle.
Understanding the Problem
- Total number of students: $n=9$.
- Condition: 2 specific students must be adjacent.
- Arrangement type: Circular path.
Calculating Total Arrangements
The total number of ways to arrange $n$ distinct items in a circle is $(n-1)!$. For 9 students, the total possible arrangements are:
$ (9-1)! = 8! $
Calculating Favorable Arrangements
To ensure 2 specific students are always together, treat them as a single unit.
- Now, we have $(9-2+1) = 8$ units to arrange in a circle (7 individual students + 1 pair).
- Number of ways to arrange these 8 units in a circle is $(8-1)! = 7!$.
- The 2 students within their unit can arrange themselves in $2!$ ways.
- Therefore, the total number of favorable arrangements is:
$ 7! \times 2! $
Determining the Probability
The probability is the ratio of favorable arrangements to total arrangements.
$ P(\text{2 students together}) = \frac{\text{Favorable Arrangements}}{\text{Total Arrangements}} $
$ P = \frac{7! \times 2!}{8!} $
Simplify the expression:
$ P = \frac{7! \times 2}{8 \times 7!} $
$ P = \frac{2}{8} $
$ P = \frac{1}{4} $
Final Answer
The probability that 2 specific students out of 9 are always standing together on a circular path is $\frac{1}{4}$.
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What is the probability that all three boys sit together?
