Solving Sitting Arrangements with and without Restrictions in a Rectangular Table
Imagine a party room with a rectangular table that seats 20 people in a row. A group of 5 individuals enters the room. How can we calculate the number of different ways for the 5 people to sit at the table both without restrictions and with the condition of having at least two empty seats between each person? This problem is a classic example in combinatorics and can be broken down into two main parts.
Part 1: Total Ways Without Restrictions
Without any restrictions, we need to determine how many ways 5 individuals can be seated out of 20 available seats.
1. Choosing Seats
The first step is to choose 5 seats out of 20. This can be calculated using the binomial formula:
[text{Number of ways} binom{n}{k} frac{n!}{k!(n-k)!}]
(n 20) (k 5)Plugging the values into the binomial formula:
[binom{20}{5} frac{20!}{5!(20-5)!} frac{20 times 19 times 18 times 17 times 16}{5 times 4 times 3 times 2 times 1} 15,504]
2. Arranging People
Once the seats are chosen, the 5 people can be seated in those 5 chosen seats in
(5! 5 times 4 times 3 times 2 times 1 120)ways.
3. Total Arrangements Without Restrictions
To find the total number of unique seating arrangements, we multiply the number of ways to choose the seats by the number of ways to arrange the people:
[text{Total} binom{20}{5} times 5! 15,504 times 120 1,860,480]
Part 2: Ways with Restrictions (At Least Two Empty Seats Between Each Person)
Now, we need to calculate the number of ways to seat the individuals with at least two empty seats between each person. This is a more complex problem that requires careful placement.
1. Placing the People
First, we place the 5 individuals. Each pair of people needs at least 2 empty seats between them. This means:
4 gaps between the 5 people, with 2 empty seats in each gap, totaling 8 empty seats Total seats occupied by people and empty seats: (5 8 13)2. Remaining Seats
Since the table has 20 seats, the remaining seats are:
[20 - 13 7] empty seats.
3. Distributing Remaining Seats
We need to distribute these 7 remaining empty seats into the 6 gaps (1 before the first person, 4 between the people, and 1 after the last person). This can be achieved using the formula for distributing identical items into distinct groups:
[binom{nk-1}{k-1}]
(n 7) (remaining seats) (k 6) (gaps)Plugging in the values:
[binom{7}{6-1} binom{12}{5} frac{12 times 11 times 10 times 9 times 8}{5 times 4 times 3 times 2 times 1} 792]
4. Arranging the People
The 5 people can still be arranged in their chosen seats in
(5! 120)5. Total Arrangements with Restrictions
To find the total number of ways the 5 people can be arranged with the given restrictions, we multiply the number of ways to distribute the empty seats by the number of ways to arrange the people:
[text{Total with restrictions} 792 times 120 95,040]
Summary of Results
Total arrangements without restrictions: 1,860,480 Total arrangements with at least two empty seats between each person: 95,040These calculations showcase the power of combinatorics and the importance of carefully analyzing the problem conditions to find the right solution.