Combinatorial Seating Arrangements and the Math Behind It

Introduction

The problem of seating 6 people in 10 seats is a classic combinatorial problem that can be solved using principles of permutations and combinations. This article will explore the mathematical reasoning behind this problem and various methods to arrive at the solution.

Choosing the Seats

To understand the number of ways to seat 6 people in 10 seats, we need to break the problem into two parts: choosing the seats and arranging the people within them.

Choosing the Seats

When choosing the seats for 6 people from 10, we are essentially selecting a combination of 6 seats out of 10. The number of ways to do this can be calculated using the combination formula:

[ binom{n}{r} frac{n!}{r!(n - r)!} ]

Here, ( n 10 ) (the total number of seats) and ( r 6 ) (the number of seats to be chosen).

Substituting the values into the formula:

[ binom{10}{6} binom{10}{4} frac{10!}{4! cdot 6!} ]

In simplified form:

[ binom{10}{6} frac{10 times 9 times 8 times 7}{4 times 3 times 2 times 1} 210 ]

This tells us there are 210 ways to choose which 6 seats will be occupied from the 10 available seats.

Arranging the People

Once the seats have been chosen, we need to consider the number of ways to arrange 6 people in those 6 selected seats. The number of possible arrangements is given by the factorial of 6 (6!):

[ 6! 720 ]

This is because there are 6 people and 6 positions, and each arrangement of these people is unique.

Total Arrangements

To find the total number of different ways to seat 6 people in 10 seats, we multiply the number of ways to choose the seats by the number of ways to arrange the people:

[ text{Total Ways} binom{10}{6} times 6! 210 times 720 151,200 ]

Thus, the total number of different ways to seat 6 people in 10 seats is 151,200.

Alternative Solutions Explained

Considering the problem from different angles can provide additional insights. Here are a few alternative solutions to the problem:

Solution 1: Treating Vacancies as Participants

In another approach, you can think of each vacant seat as a participant. With 9 participants (7 people and 2 vacancies), the number of ways to arrange them is 9! (factorial of 9). However, since the two vacancies are interchangeable, they do not add to the diversity of arrangements. Therefore, you must divide the total arrangements by 2:

[ frac{9!}{2} frac{362,880}{2} 181,440 ]

This is a valid approach but slightly overcounts the total arrangements.

Solution 2: Fixing the Vacancies

Another way to solve the problem is to first select 2 seats that will be left empty. The number of ways to choose 2 empty seats from 9 is the combination formula for choosing 2 from 9:

[ binom{9}{2} frac{9!}{2!(9-2)!} frac{9!}{2!7!} ]

Upon selecting the 2 empty seats, you have 7 seats left for 7 people to arrange, which is given by 7! (factorial of 7).

The product of these two values gives the total number of arrangements:

[ binom{9}{2} times 7! frac{9!}{2!7!} times 7! frac{9!}{2!} 181,440 ]

This solution also leads to 181,440 arrangements, which is essentially an overcounted version of the initial solution.

Conclusion

The initial solution correctly calculates the total number of different ways to seat 6 people in 10 seats as 151,200. While alternative solutions provide useful insights, they can sometimes overcount the arrangements. The essential takeaway is that the combination of selecting seats and arranging people yields the precise number of unique seating arrangements.