An electronics store receives a shipment of 20 graphing calculators, including 3 that are defective. Four of the calculators are selected to be sent to a local high school. A. How many selection can be made using the originial shipment?

An electronics store receives a shipment of 20 graphing calculators, including 2 that are defective. Four of the calculators are selected to be sent to a local high school. A. How many selection can be made using the originial shipment?

Well, if we're talking about the original shipment of 20 calculators, then we can use basic combinatorics to find the number of selections. Since we're selecting 4 calculators from a set of 20, we can use the combination formula, which is nCr = n! / (r!(n-r)!). In this case, n = 20 (because there are 20 calculators in total) and r = 4 (because we're selecting 4 calculators).

So the number of selections can be calculated as follows:

20C4 = 20! / (4!(20-4)!)
= 20! / (4!16!)
= (20 * 19 * 18 * 17) / (4 * 3 * 2 * 1)
= 4845.

Therefore, there are 4845 possible selections that can be made using the original shipment of 20 calculators.

To determine the number of selections that can be made using the original shipment of 20 graphing calculators, we can use the concept of combinations.

The formula for calculating combinations is represented as "nCr," where n is the total number of items and r is the number of items to be selected from the total.

In this case, we have 20 graphing calculators, and we want to select 4 of them. So, we can calculate the number of selections using 20C4.

The formula for nCr is:

nCr = n! / [(n-r)! * r!]

First, let's calculate n! (factorial of n):
20! = 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

Next, let's calculate (n-r)!
(20-4)! = 16! = 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

Lastly, let's calculate r!
4! = 4 x 3 x 2 x 1

Now, substitute the values into the formula for nCr:

20C4 = 20! / [(20-4)! * 4!]
= (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) / [(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) * (4 x 3 x 2 x 1)]

Calculating this expression will give us the total number of selections that can be made using the original shipment of graphing calculators.

If the only distinction between calculators is the defects then there are 4 possible selections, namely

0, 1, 2, 3 defective calculators along with good ones for the balance.

If the calculators bear serial numbers, i.e. they are considered distinct, then there are (20,4)=20!/(4!16!)=4845 selections.

An electronics store receives a shipment of 20 graphing​ calculators, including 4 that are defective. Four of the calculators are selected to be sent to a local high school. How many of these selections will contain no defective​ calculators?