A group of 9 students s₁, s₂, ..., s₉ is to be divided to form three teams X, Y and Z of sizes 2, 3

Q: Why can't we just calculate the total ways to form teams and then subtract the cases where $s_1 \in X$ or $s_2 \in Y$?

That is a great intuition, but it leads to the Principle of Inclusion-Exclusion, which can get messy here. You would have to calculate the total ways, subtract the cases where $s_1 \in X$, subtract the cases where $s_2 \in Y$, and then add back the intersection where both $s_1 \in X$ and $s_2 \in Y$. It is much cleaner and less prone to error to simply partition the problem into the 4 mutually exclusive cases we identified.

Q: How do I know if I have missed a case?

The best way to verify is to list the possible homes for the restricted students. $s_1$ can be in $Y$ or $Z$. $s_2$ can be in $X$ or $Z$. The Cartesian product of these sets of options gives all combinations: $(Y, X), (Y, Z), (Z, X), (Z, Z)$. Since these cover all valid possibilities for $s_1$ and $s_2$, you have accounted for every scenario.

Q: Does the order of students within the teams matter?

No, the problem asks for the number of ways to form teams. In team formation, the order of members within a team does not matter; only the membership itself matters. That is why we use the combination formula $\binom{n}{r}$ rather than permutations.

Question

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 ________.

Sigma Percentile · Accepted Answer

### The Art of Strategic Counting

Imagine you are the captain of a grand tournament, tasked with organizing nine brilliant students into three distinct teams: Team $X$ (size 2), Team $Y$ (size 3), and Team $Z$ (size 4).

It sounds simple, right? But life, much like a JEE Advanced problem, is rarely without its constraints.

We have two particular students, $s_1$ and $s_2$, who have specific demands. Student $s_1$ refuses to join Team $X$, and student $s_2$ will not step foot in Team $Y$. Our mission is to find the total number of ways to form these teams while respecting their wishes.

### Phase 1: The Power of Partitioning

When faced with complex constraints, the most powerful tool in your arsenal is to break the problem into mutually exclusive cases. Instead of panicking about the constraints, let us embrace them.

Since $s_1$ cannot be in Team $X$, they must be in either Team $Y$ or Team $Z$. Similarly, since $s_2$ cannot be in Team $Y$, they must be in either Team $X$ or Team $Z$.

By looking at these as independent choices, we naturally arrive at four distinct, non-overlapping scenarios:

1. $s_1 \in Y$ and $s_2 \in X$ 
2. $s_1 \in Z$ and $s_2 \in X$ 
3. $s_1 \in Y$ and $s_2 \in Z$ 
4. $s_1 \in Z$ and $s_2 \in Z$

### Phase 2: Solving the Scenarios

Let us tackle these one by one. In every case, we have 9 students total, and after placing $s_1$ and $s_2$, we have 7 students remaining to fill the remaining slots.

**Case 1: $s_1 \in Y$ and $s_2 \in X$**

Here, Team $X$ needs 1 more student (out of 7), Team $Y$ needs 2 more, and Team $Z$ needs 4. The number of ways is:

$$ \binom{7}{1} 	imes \binom{6}{2} 	imes \binom{4}{4} = 7 	imes 15 	imes 1 = 105 $$

**Case 2: $s_1 \in Z$ and $s_2 \in X$**

Team $X$ needs 1 more, Team $Y$ needs 3, and Team $Z$ needs 3. The calculation is:

$$ \binom{7}{1} 	imes \binom{6}{3} 	imes \binom{3}{3} = 7 	imes 20 	imes 1 = 140 $$

**Case 3: $s_1 \in Y$ and $s_2 \in Z$**

Team $X$ needs 2, Team $Y$ needs 2, and Team $Z$ needs 3. The calculation is:

$$ \binom{7}{2} 	imes \binom{5}{2} 	imes \binom{3}{3} = 21 	imes 10 	imes 1 = 210 $$

**Case 4: $s_1 \in Z$ and $s_2 \in Z$**

Team $X$ needs 2, Team $Y$ needs 3, and Team $Z$ needs 2. The calculation is:

$$ \binom{7}{2} 	imes \binom{5}{3} 	imes \binom{2}{2} = 21 	imes 10 	imes 1 = 210 $$

### Phase 3: The Grand Total

Now, we simply sum these mutually exclusive cases to find our answer:

$$ 105 + 140 + 210 + 210 = 665 $$

It is elegant, it is logical, and it is complete. By breaking down the problem, we turned a daunting constraint into a clear, manageable path.

Keep this strategy in mind: whenever you see a constraint, don't just see a restriction—see a way to partition your problem into simpler, solvable pieces. You have got this! The final answer is **665**.

Enter Numerical Value:

Visualized Solution

The Teams and Students

The Constraints

Possible Placements

Case 1: $s_{1} \in Y$ and $s_{2} \in X$

Calculating Case 1

Case 2: $s_{1} \in Z$ and $s_{2} \in X$

Calculating Case 2

Case 3: $s_{1} \in Y$ and $s_{2} \in Z$

Calculating Case 3

Case 4: $s_{1} \in Z$ and $s_{2} \in Z$

Calculating Case 4

Total Number of Ways

The Sigma Insight: Formation of Groups

The Art of Strategic Counting

Phase 1

Phase 2

Phase 3

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