In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such

Q: Wait, why don't we divide by $2!$ when we distribute the people into the rooms with capacity $2$?

This is a classic trap! We only divide by the factorial of the group sizes if the groups themselves are indistinguishable. However, in this problem, the rooms are distinct (e.g., Room $101$ is different from Room $102$). Once we have assigned the capacities to specific rooms, the rooms act as unique labels. Therefore, we do not divide by $2!$ for the rooms; the distinctness of the rooms already accounts for the arrangement.

Q: Can we solve this using generating functions?

Yes, you absolutely can! You would look for the coefficient of $x^6$ in the expansion of $(x + \frac{x^2}{2!})^4 \cdot 4!$. However, for a problem of this size, the combinatorial approach of partitioning and multiplying is much faster and less prone to algebraic errors. Stick to the two-phase method: assign capacities, then distribute.

Q: What if the rooms were identical? Would the answer change?

Great question! If the rooms were identical, the order in which we assign the capacities wouldn't matter, and we would indeed have to divide by the symmetry of the identical rooms (the $2!$ for the two rooms with $2$ people and the $2!$ for the two rooms with $1$ person). But since the problem specifies a 'hotel' with 'four rooms', we assume they are distinct entities.

Question

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 ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Let $x_i$ be the number of persons in room $i$, where $i \in \{1, 2, 3, 4\}$. We are given two primary constraints: the total number of people is $x_1 + x_2 + x_3 + x_4 = 6$, and the capacity constraint is $1 \le x_i \le 2$.

We must partition the number $6$ into $4$ parts, where each part is either $1$ or $2$. The only mathematical solution to satisfy this sum is:

$$2 + 2 + 1 + 1 = 6$$

This implies that exactly $2$ rooms will be double-occupancy, and $2$ rooms will be single-occupancy.

### Assigning Capacities to Rooms

Since the rooms are distinct, we must determine which rooms receive which capacity. We need to select $2$ rooms out of $4$ to have a capacity of $2$.

The number of ways to choose these rooms is given by the combination:

$${}^4C_2 = \frac{4!}{2! \cdot 2!} = 6 	ext{ ways}$$

We have now fixed the structural configuration of the hotel occupancy.

### Distributing the Guests

With the capacities fixed, we distribute our $6$ distinct guests into these specific slots. The number of ways to partition $6$ distinct people into groups of sizes $2, 2, 1, 1$ is given by the multinomial coefficient:

$$\frac{6!}{2! \cdot 2! \cdot 1! \cdot 1!} = \frac{720}{4} = 180 	ext{ ways}$$

A common pitfall is to divide by $2!$ to account for the identical group sizes. However, because the rooms are distinct, the groups are already distinguished by the specific rooms they occupy.

We have already accounted for the room identities in the previous phase. Therefore, $180$ is the correct number of ways to distribute the people for any one of the $6$ capacity configurations.

### The Grand Finale

To find the total number of ways to accommodate the guests, we multiply the number of ways to assign the capacities by the number of ways to distribute the people:

$$6 	imes 180 = 1080$$

The total number of ways to arrange the guests is **1080**.

Enter Numerical Value:

Visualized Solution

$4$ Rooms, $6$ Persons

Constraints: $1 \leq x_{i} \leq 2$

Possible Capacities: ${2, 2, 1, 1}$

Assigning Capacities to Rooms

Distributing $6$ Persons

Total Number of Ways

Conclusion

The Sigma Insight: Formation of Groups

Analyzing the Setup

Assigning Capacities to Rooms

Distributing the Guests

The Grand Finale

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4 Rooms, .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6 Persons

Constraints: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1≤xi​≤2

Possible Capacities: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } {2,2,1,1}

Assigning Capacities to Rooms

Distributing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6 Persons

Total Number of Ways

Conclusion

The Sigma Insight: Formation of Groups

Analyzing the Setup

Assigning Capacities to Rooms

Distributing the Guests

The Grand Finale

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The set S={1,2,3,…,12} is to be partitioned into three sets A,B,C of equal size. Thus A∪B∪C=S,A∩B=B∩C=A∩C=ϕ. The number of ways to partition S is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α=(4!)3!(4!)!​ and β=(5!)4!(5!)!​. Then :

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