Eighteen guests have to be seated, half on each side of a long table. Four particular guests desire

Q: Wait, why do we multiply by $9!$ and $9!$ at the end? Shouldn't we just select the groups?

Great question! Combinatorics is a two-step dance. Selecting the group (using combinations) only tells us *who* is sitting on which side. But the problem asks for the number of seating arrangements, which implies that the order of the guests in their chairs matters. Since there are $9$ distinct chairs on each side, we must arrange the $9$ people on Side A in $9!$ ways and the $9$ people on Side B in $9!$ ways.

Q: Does it matter if I choose the guests for Side B first instead of Side A?

Not at all! If you choose $6$ guests for Side B from the $11$ available, you would use ${}^{11}C_6$. Mathematically, ${}^{11}C_6$ is exactly the same as ${}^{11}C_5$. The result remains identical because once you fill one side, the other side is automatically determined by the remaining guests.

Q: Why can't we just use ${}^{18}C_9$ to pick the people for one side?

You would use ${}^{18}C_9$ if there were no restrictions. However, because $4$ guests are already locked into Side A and $3$ are locked into Side B, you don't have $18$ free choices. You only have $11$ free guests to distribute. The constraints drastically reduce the sample space, so we must account for the fixed guests first.

Question

Eighteen guests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side. Determine the number of ways in which the sitting arrangements can be made.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with seating $18$ guests at a long table with $9$ seats on each side. We must respect the specific preferences of $7$ guests: $4$ guests must sit on Side A, and $3$ guests must sit on Side B.

This leaves us with $18 - 7 = 11$ flexible guests who have no specific seating preference.

### The Selection Process

Side A has $9$ total seats, but $4$ are already occupied by the fixed guests. This leaves $9 - 4 = 5$ empty seats to be filled on Side A.

Side B has $9$ total seats, with $3$ occupied by the fixed guests. This leaves $9 - 3 = 6$ empty seats to be filled on Side B.

We must choose $5$ guests from the $11$ flexible individuals to occupy the remaining seats on Side A. The number of ways to perform this selection is given by the combination formula:

$${}^{11}C_{5} = \frac{11!}{5!6!}$$

Once these $5$ guests are chosen, the remaining $6$ flexible guests are automatically assigned to the $6$ empty seats on Side B. The number of ways to choose $6$ from $6$ is simply ${}^{6}C_{6} = 1$.

### The Arrangement

Now that the groups are assigned to their respective sides, we must account for the permutations of the individuals within those seats.

On Side A, we have a total of $9$ distinct guests (the $4$ fixed plus the $5$ selected). These guests can be arranged in the $9$ chairs in $9!$ ways.

On Side B, we also have a total of $9$ distinct guests (the $3$ fixed plus the $6$ remaining). These guests can be arranged in the $9$ chairs in $9!$ ways.

### The Final Calculation

By the Fundamental Principle of Counting, we multiply the number of ways to select the groups by the number of ways to arrange the guests on each side.

The total number of valid arrangements is:

$${}^{11}C_{5} 	imes 9! 	imes 9!$$

Substituting the value of the combination, the final result is:

$$\frac{11!}{5!6!} 	imes 9! 	imes 9!$$

This expression represents the total number of ways to seat the guests while satisfying all given constraints.

Visualized Solution

Table Setup

Guest Preferences

Remaining Guests

Available Slots

Selecting for Side A

Selecting for Side B

Arranging Side A

Arranging Side B

Total Arrangements

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Selection Process

The Arrangement

The Final Calculation

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