Two different families A and B are blessed with equal number of children. There are 3 tickets to be

Q: Wait, why do we use combinations $\binom{n}{r}$ instead of permutations $P(n,r)$?

Great question! In this problem, we are simply distributing 3 tickets to 3 different children. The problem does not state that the tickets are distinct (e.g., Ticket 1, Ticket 2, Ticket 3). Even if the tickets were distinct, the probability would remain the same because the number of ways to arrange the tickets among the selected children would appear in both the numerator and the denominator, canceling each other out. Thus, we treat this as a selection problem.

Q: Why is the denominator $\binom{2n}{3}$ and not something else?

The denominator represents the total number of ways to choose 3 children out of the entire group. Since Family A has $n$ children and Family B has $n$ children, the total number of children is $n + n = 2n$. We are choosing 3 children from this total pool of $2n$ children, hence $\binom{2n}{3}$.

Q: Can $n$ be less than 3? What if the math gives me a smaller value?

Mathematically, for the combination $\binom{n}{3}$ to be defined in the context of selecting 3 children, $n$ must be at least 3. If you were to get a value like $n=2$, it would be physically impossible to choose 3 children from that family. Always check your final answer against the physical constraints of the problem!

Question

Two different families A and B are blessed with equal number of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is $\frac{1}{12}$, then the number of children in each family is :

Options:

(A) 3

(B) 5

(C) 4

(D) 6

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a room filled with children from two different families, Family A and Family B. Both families are blessed with an equal number of children, let us call this number $n$.

The air is thick with anticipation because there are 3 tickets to be distributed. The rule is simple yet strict: no child can receive more than one ticket.

### Visualizing the Selection

When we talk about selecting 3 children from a total pool of $2n$ children (since $n$ from A plus $n$ from B equals $2n$), we are entering the realm of combinations. We are not concerned with the order of the tickets; we only care about who gets them.

The total number of ways to choose 3 children from the entire group of $2n$ is given by the combination formula $\binom{2n}{3}$. This is our sample space, the denominator of our probability fraction.

### The Favorable Outcome

Now, consider the specific event mentioned in the problem: all 3 tickets go to the children of Family B. To make this happen, we must select all 3 children exclusively from the $n$ children of Family B.

The number of ways to do this is $\binom{n}{3}$. This is our favorable outcome. The probability $P$ of this event is the ratio of favorable outcomes to total outcomes:

$$P = \frac{\binom{n}{3}}{\binom{2n}{3}}$$

We are told that this probability is exactly $\frac{1}{12}$.

### The Algebraic Dance

Now, let us dive into the algebra. We know that $\binom{k}{3} = \frac{k(k-1)(k-2)}{6}$. Substituting this into our probability equation, we get:

$$\frac{\frac{n(n-1)(n-2)}{6}}{\frac{2n(2n-1)(2n-2)}{6}} = \frac{1}{12}$$

Notice the beauty of this step: the $6$ in the denominator of both the numerator and the denominator cancels out perfectly! We are left with:

$$\frac{n(n-1)(n-2)}{2n(2n-1)(2n-2)} = \frac{1}{12}$$

### The Final Simplification

Look closely at the term $(2n-2)$ in the denominator. We can factor out a $2$ to get $2(n-1)$. Our equation now looks like this:

$$\frac{n(n-1)(n-2)}{2n(2n-1) \cdot 2(n-1)} = \frac{1}{12}$$

Now, we can cancel the common terms $n$ and $(n-1)$ from the top and bottom. This leaves us with:

$$\frac{n-2}{4(2n-1)} = \frac{1}{12}$$

Multiplying both sides by $4$, we get $\frac{n-2}{2n-1} = \frac{4}{12} = \frac{1}{3}$. Cross-multiplying gives us $3(n-2) = 1(2n-1)$, which simplifies to $3n - 6 = 2n - 1$.

Solving for $n$, we find **$n = 5$**. Each family has 5 children. It is an elegant result, born from a simple, logical journey through the world of combinations.

Select Answer:

Visualized Solution

Visualizing the Families

The Tickets and The Constraint

Total Possible Outcomes

Favorable Outcomes for Family B

Setting up the Probability Equation

Expanding the Combinations

Simplifying the Expression

Canceling Common Terms

Solving for $n$

Final Conclusion

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

Visualizing the Selection

The Favorable Outcome

The Algebraic Dance

The Final Simplification

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