An electronic store receives a shipment of 30 graphing calculators, including 6 that are defective. Four of these calculators are selected to be sent to a local high school. How many selections can be made? How many of this selections will contain two defective calculators?
30 C 4
To determine the total number of selections that can be made, we need to use the concept of combinations. The formula for combinations is as follows:
C(n, r) = n! / (r!(n - r)!)
Where:
C(n, r) is the number of combinations of n items taken r at a time.
n is the total number of items.
r is the number of items to be selected.
! represents the factorial operation (e.g., n! = n * (n-1) * (n-2) * ... * 1).
In this case, there are 30 graphing calculators, and we want to select 4 of them.
The number of selections that can be made is given by:
C(30, 4) = 30! / (4!(30 - 4)!)
Calculating this, we have:
C(30, 4) = 30! / (4! * 26!)
= (30 * 29 * 28 * 27 * 26!) / (4! * 26!)
= (30 * 29 * 28 * 27) / (4 * 3 * 2 * 1)
= 27,405
So, there are 27,405 possible selections that can be made.
Now, let's determine the number of selections that will contain two defective calculators.
We have 6 defective calculators out of a total of 30. To select 2 defective calculators, we can use combinations again.
The number of combinations of selecting 2 defective calculators out of 6 is given by:
C(6, 2) = 6! / (2!(6 - 2)!)
Calculating this, we have:
C(6, 2) = 6! / (2! * 4!)
= (6 * 5 * 4!) / (2 * 1 * 4!)
= (6 * 5) / (2 * 1)
= 15
So, there are 15 selections that will contain two defective calculators.
To find the number of selections that can be made, we need to use the concept of combinations.
The total number of selections that can be made from a set of 30 calculators is given by the combination formula:
nCr = n! / (r!(n-r)!)
where n is the total number of items in the set, r is the number of items to be selected, and ! represents the factorial function.
In this case, we have 30 calculators and we want to select 4 of them. So, plugging in the given values into the formula:
30C4 = 30! / (4!(30-4)!)
= 30! / (4!26!)
Now, calculating the factorial values:
30! = 30 x 29 x 28 x 27 x 26!
4! = 4 x 3 x 2 x 1 = 24
26! = 26 x 25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 403291461126605650322784256!
Plugging in the values:
30C4 = (30 x 29 x 28 x 27 x 26!) / (24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1)
Using a calculator or computer program to simplify the expression, we find that 30C4 = 27,405 possible selections can be made from the 30 graphing calculators.
To find the number of selections that contain two defective calculators, we can use the combination formula again with the given information. In this case, we have 6 defective calculators and we want to select 2 of them. Therefore:
6C2 = 6! / (2!(6-2)!)
= 6! / (2!4!)
Calculating the factorial values:
6! = 6 x 5 x 4!
2! = 2 x 1 = 2
4! = 4 x 3 x 2 x 1 = 24
Plugging in the values:
6C2 = (6 x 5 x 4!) / (2 x 1 x 4!)
Simplifying the expression, we find that 6C2 = 15 selections will contain two defective calculators.
a) 30C4=27405
b) 6C2x24C2=4140