Key Concepts and Formulas

1. Combinations: The number of ways to choose $k$ distinct items from a set of $n$ distinct items, without regard to the order of selection, is given by the combination formula: ${}^n C_k = \binom{n}{k} = \frac{n!}{k!(n-k)!}$ For practical calculations, this can be expanded as $\frac{n(n-1)(n-2)...(n-k+1)}{k!}$.
2. Probability: The probability of an event occurring is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes, assuming all outcomes are equally likely: $P(\text{Event}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$

Step-by-Step Solution

Let $n$ be the number of children in each family (Family A and Family B).

• Number of children in Family A = $n$
• Number of children in Family B = $n$
• Total number of children = $n + n = 2n$
• Number of tickets to be distributed = 3

The condition "no child gets more than one ticket" means we are selecting 3 distinct children from the total pool. Since the tickets are identical in purpose, the order of selection does not matter, so we use combinations. Also, for combinations ${}^n C_k$ to be defined, $n \ge k$. In this case, $n \ge 3$.

Step 1: Calculate Total Possible Outcomes

• What we are doing: We need to find the total number of ways to distribute 3 tickets among the $2n$ children. This means choosing 3 children out of $2n$ available children.
• Why: This forms the denominator of our probability fraction.
• Calculation: Using the combination formula, the total number of ways to choose 3 children from $2n$ children is: ${}^{2n} C_3 = \frac{(2n)!}{3!(2n-3)!} = \frac{2n \times (2n-1) \times (2n-2)}{3 \times 2 \times 1}$ ${}^{2n} C_3 = \frac{2n(2n-1)(2n-2)}{6}$

Step 2: Calculate Favorable Outcomes

• What we are doing: We are interested in the event where all 3 tickets go exclusively to the children of Family B. This means we need to choose 3 children solely from the $n$ children belonging to Family B.
• Why: This forms the numerator of our probability fraction.
• Calculation: The number of ways to choose 3 children from the $n$ children in Family B is: ${}^n C_3 = \frac{n!}{3!(n-3)!} = \frac{n \times (n-1) \times (n-2)}{3 \times 2 \times 1}$ ${}^n C_3 = \frac{n(n-1)(n-2)}{6}$

Step 3: Formulate the Probability Equation

• What we are doing: We use the given probability, $1/12$, and our calculated total and favorable outcomes to set up an equation.
• Why: This equation will allow us to solve for $n$.
• Equation: The problem states that the probability of all tickets going to Family B is $\frac{1}{12}$. $P(\text{all tickets to Family B}) = \frac{{}^n C_3}{{}^{2n} C_3} = \frac{1}{12}$ Substitute the expressions for ${}^n C_3$ and ${}^{2n} C_3$: $\frac{\frac{n(n-1)(n-2)}{6}}{\frac{2n(2n-1)(2n-2)}{6}} = \frac{1}{12}$

Step 4: Solve the Equation for 'n'

• What we are doing: We simplify the algebraic equation and solve for the variable $n$.
• Why: $n$ represents the number of children in each family, which is what the question asks for.
• Calculation: First, cancel out the common denominator of 6 from the numerator and denominator of the complex fraction: $\frac{n(n-1)(n-2)}{2n(2n-1)(2n-2)} = \frac{1}{12}$ Since $n \ge 3$, $n$, $n-1$, and $n-2$ are all non-zero. We can simplify by canceling common terms: $\frac{n(n-1)(n-2)}{2n(2n-1) \cdot 2(n-1)} = \frac{1}{12}$ Cancel $n$ and $(n-1)$ from the numerator and denominator: $\frac{n-2}{2(2n-1) \cdot 2} = \frac{1}{12}$ $\frac{n-2}{4(2n-1)} = \frac{1}{12}$ Now, cross-multiply: $12(n-2) = 4(2n-1)$ Divide both sides by 4 to simplify: $3(n-2) = 2n-1$ Distribute the 3: $3n - 6 = 2n - 1$ Gather terms involving $n$ on one side and constant terms on the other: $3n - 2n = 6 - 1$ $n = 5$

Verification: If $n=5$:

• Number of children in Family B = 5.
• Total number of children = $2 \times 5 = 10$.
• Favorable outcomes: ${}^5 C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$.
• Total possible outcomes: ${}^{10} C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
• Probability = $\frac{10}{120} = \frac{1}{12}$. This matches the given probability.

Common Mistakes & Tips

• Combinations vs. Permutations: Always determine if the order of selection matters. "No child gets more than one ticket" and "distribute tickets" typically indicate combinations.
• Constraints on $n$: Remember that for ${}^n C_k$ to be valid, $n$ must be greater than or equal to $k$. Here, $n \ge 3$, which justifies cancelling terms like $n$, $n-1$, and $n-2$.
• Algebraic Simplification: Be careful and methodical with algebraic manipulations. Factoring out common terms (e.g., $2n-2 = 2(n-1)$) can prevent errors.
• Verification: Always plug your final answer back into the original probability equation to ensure it satisfies the given conditions.

Summary

This problem involved calculating the probability of a specific event using combinations. By defining the number of children in each family as $n$, we determined the total number of ways to distribute 3 tickets among $2n$ children (${}^{2n} C_3$) and the number of ways all 3 tickets could go to Family B (${}^n C_3$). Setting the ratio of these two values equal to the given probability of $\frac{1}{12}$ led to an algebraic equation. Solving this equation correctly yields $n=5$.

The final answer is $\boxed{\text{3}}$ which corresponds to option (A).