The students S1, S2, ..., S10 are to be divided into 3 groups A, B and C such that each group has at

Q: Why do we subtract 2 in the formula $2^k - 2$?

Great question! When we distribute $k$ distinct students into two distinct rooms (A and B), each student has 2 choices: room A or room B. This gives us $2^k$ total possible assignments. However, this count includes two invalid scenarios: one where all students end up in room A (leaving B empty) and one where all students end up in room B (leaving A empty). Since the problem explicitly states that each group must have at least one student, we must subtract these 2 invalid cases.

Q: Does the order of groups A, B, and C matter?

Yes, it absolutely does. The problem defines the groups as A, B, and C. In combinatorics, if the groups were indistinguishable, we would have to divide by the number of permutations of the groups. But here, the groups are distinct entities (like three different rooms), so we treat them as labeled and distinct.

Q: Can we use the Principle of Inclusion-Exclusion for this?

You certainly could, but it would be significantly more complex than the case-by-case method. Inclusion-Exclusion is powerful, but when constraints are as simple as 'group C has at most 3', breaking the problem into mutually exclusive cases is the most elegant and error-proof path to the solution.

Question

The students $S_1, S_2, \dots, S_{10}$ are to be divided into 3 groups $A, B$ and $C$ such that each group has at least one student and the group $C$ has at most 3 students. Then the total number of possibilities of forming such groups is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a large hall with ten brilliant students, and your task is to organize them into three distinct rooms: Room A, Room B, and Room C. This is a puzzle of constraints where every room must have at least one student, and Room C can hold at most three students.

The secret to solving this lies in the power of systematic decomposition.

### The Strategy of Cases

When faced with a constraint like "at most 3," the best strategy is to break the problem into smaller, manageable pieces. We define our cases based on the number of students in Room C, denoted as $n_C$.

Since $n_C$ must be at least 1 and at most 3, we have exactly three mutually exclusive scenarios: $n_C = 1$, $n_C = 2$, and $n_C = 3$. By solving each case independently, we ensure we do not double-count or miss any possibilities.

### The Core Logic: The $2^k - 2$ Rule

Before we calculate, let us establish our primary tool. Suppose we have $k$ students left to place into two distinct rooms, A and B.

Each student has two choices, leading to $2^k$ total ways. However, we cannot leave a room empty. The only invalid ways are when all $k$ students are in A (leaving B empty) or all $k$ students are in B (leaving A empty).

Thus, the number of valid ways to fill two non-empty rooms is given by the formula:
$$N = 2^k - 2$$

### Act I: The Case of One

For $n_C = 1$, we first select one student for Room C in $\binom{10}{1}$ ways. We are left with 9 students to place into A and B.

Using our rule, the number of ways to distribute the remaining students is:
$$2^9 - 2 = 512 - 2 = 510$$

Therefore, for Case 1, the total number of ways is:
$$10 	imes 510 = 5100$$

### Act II: The Case of Two

For $n_C = 2$, we choose two students for Room C in $\binom{10}{2}$ ways, which equals 45. We have 8 students remaining for rooms A and B.

Applying the rule, we get:
$$2^8 - 2 = 256 - 2 = 254$$

Multiplying these values, we find the total for Case 2:
$$45 	imes 254 = 11430$$

### Act III: The Case of Three

For $n_C = 3$, we select three students for Room C in $\binom{10}{3}$ ways, which equals 120. We have 7 students left for rooms A and B.

Applying the rule, we get:
$$2^7 - 2 = 128 - 2 = 126$$

Multiplying these values, we find the total for Case 3:
$$120 	imes 126 = 15120$$

### The Grand Finale

Since these cases are mutually exclusive, we simply add them together to find the total number of arrangements:
$$5100 + 11430 + 15120 = 31650$$

By breaking the problem down, we turned a daunting task into a series of simple, satisfying calculations. The final answer is **31650**.

Enter Numerical Value:

Visualized Solution

Problem Setup

The Constraints

Defining the Cases

Distribution Formula

Case 1: $n_{C} = 1$

Case 1 Calculation

Case 2: $n_{C} = 2$

Case 2 Calculation

Case 3: $n_{C} = 3$

Case 3 Calculation

Total Possibilities

The Sigma Insight: Formation of Groups

Analyzing the Setup

The Strategy of Cases

The Core Logic

Act I

Act II

Act III

The Grand Finale

Similar Questions

A group of 9 students $s_{1}, s_{2}, \dots, s_{9}$ is to be divided to form three teams $X, Y$ and $Z$ of sizes 2, 3 and 4 respectively. Suppose that $s_{1}$ cannot be selected for the team $X$ , and $s_{2}$ cannot be selected for team $Y$ . Then the number of ways to form such teams, is ________.

The set $S = {1, 2, 3, \dots, 12}$ is to be partitioned into three sets $A, B, C$ of equal size. Thus $A \cup B \cup C = S, A \cap B = B \cap C = A \cap C = ϕ$ . The number of ways to partition $S$ is

Eight persons are to be transported from city A to city B in three cars of different makes. If each car can accommodate at most three persons, then the number of ways, in which they can be transported, is

In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons. Then the number of all possible ways in which this can be done is ____.

Let the set $S = {2, 4, 8, 16, \dots, 512}$ be partitioned into 3 sets $A, B, C$ with equal number of elements such that $A \cup B \cup C = S$ and $A \cap B = B \cap C = A \cap C = ϕ$ . The maximum number of such possible partitions of $S$ is equal to:

If $n$ is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then $n$ is equal to:

Let $X$ be a set with exactly $5$ elements and $Y$ be a set with exactly $7$ elements. If $α$ is the number of one-one functions from $X$ to $Y$ and $β$ is the number of onto functions from $Y$ to $X$ , then the value of $\frac{1}{5 !} (β - α)$ is ________.

Let $α = \frac{( 4 !)!}{( 4 ! ) ^{3!}}$ and $β = \frac{( 5 !)!}{( 5 ! ) ^{4!}}$ . Then :