An astronomer wishes to measure on a photograph the distance between the image of a certain star and three other star images nearby. If eight images are nearby, how many choices of the three stars does he have?
Number of choices: 8C3 = 56
An astronomer wishes to measure on a photograph the distance between the image of a certain star and three other star images nearby. If eight images are nearby, how many choices of the three stars does he have?
To determine the number of choices the astronomer has to measure the distance between the image of the certain star and three other star images nearby, we can use the concept of combinations.
The number of choices of k objects from a set of n objects is given by the formula for combinations, which can be expressed as:
C(n, k) = n! / (k! * (n - k)!),
where n! denotes the factorial of n.
In this case, the astronomer has eight star images nearby and wants to choose three stars. Hence, the number of choices can be calculated as:
C(8, 3) = 8! / (3! * (8 - 3)!) = 8! / (3! * 5!)
Simplifying this expression, we get:
C(8, 3) = (8 * 7 * 6 * 5!) / (3! * 5!) = (8 * 7 * 6) / (3 * 2 * 1) = 56.
Therefore, the astronomer has a total of 56 choices to measure the distance between the image of a certain star and three other star images nearby.
To determine the number of choices the astronomer has, we need to consider the concept of combinations.
In this scenario, the astronomer wants to choose three star images out of the eight nearby images. The order in which the stars are chosen does not matter; what matters is the combination of the stars.
The formula to calculate combinations is given by:
C(n, k) = n! / (k! * (n-k)!)
Where:
- n is the total number of objects (eight star images in this case)
- k is the number of objects to be chosen (three star images in this case)
- ! represents the factorial operator (e.g., 4! is 4 factorial, which is 4 * 3 * 2 * 1)
Plugging in the values into the formula, we have:
C(8, 3) = 8! / (3! * (8-3)!)
= 8! / (3! * 5!)
Now let's calculate the factorial values:
8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
3! = 3 * 2 * 1
5! = 5 * 4 * 3 * 2 * 1
Substituting these values into the formula:
C(8, 3) = (8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1 * (5 * 4 * 3 * 2 * 1))
Upon simplifying the expression, we get:
C(8, 3) = 8 * 7 * 6 / (3 * 2 * 1) = 56
Therefore, the astronomer has 56 choices of three stars among the eight nearby star images.