six people are going to be seated in a row of 8 chairs. how many different ways may they be seated?
Pretend you have two extra people to fill the empty seats,
Then the number of ways to arrange them would be 8!
But the two really were not there, and the two empty seats are like having "2 like things"
so final number of arrangements = 8!/2! = 20160
To find the number of different ways that six people can be seated in a row of 8 chairs, we can use the concept of permutations.
The number of ways to arrange six people in a row of eight chairs can be calculated using the formula for permutations. In this case, we need to consider that the order of seating matters (i.e., the seating arrangements are distinct).
The formula for permutations is given by:
P(n,r) = n! / (n - r)!
Where:
- P(n,r) is the number of permutations of selecting r objects from a set of n objects.
- n! denotes the factorial of n, which means multiplying all the integers from 1 to n.
In this case, we have 8 chairs to choose from and 6 people to seat. Therefore, the formula becomes:
P(8,6) = 8! / (8 - 6)!
= 8! / 2!
Now, let's calculate the permutations:
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320
2! = 2 x 1 = 2
Now we can substitute the values back into the formula:
P(8,6) = 40,320 / 2
= 20,160
Therefore, there are 20,160 different ways the six people can be seated in a row of eight chairs.