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Permutations and Combinations


In the first video, we are going to work with permutations.

We have three cars, the blue car, the white car, and the green car.
We can have different orders of:
B W G,
B G W,
W B G,
W G B,
G B W, and
which adds up to the total number of order of 6 possibilities.

Another way to look at this this would be thinking the cars in order:

  • In the first place, there can be three cars.
  • In the second place, there can be only two cars since one car is already in the first place.
  • In the third place, there can be only one car since there is only one left.

Multiplying 3 by 2 by 1, we have 6. which is the same answer as above.

We can simply use factorial, “!” to solve this kind of questions.

For 4!, the expression can also be written as
4\times3\times2\times1 from the given number to 1

Multiplying the numbers together and we have

Let’s work on another example:
If there are 5 cars, in how many ways can they finish for the first three cars?

In this situation, we cannot directly use 5! because we are not finding the order for all five cars.
Instead, we can apply multiplication up to the desired order, to which we call permutations. We use permutations in situations where the order and position matters.


On a calculator, we can type in the first number, press MATH/PRB nPr, and enter the second number. The answer is the same in both situations.

Permuations with Repeated Elements

In the second video, we are going to work with permutations with repeated elements.

For the word KNEE, there are three distinctive letters with the letter E being repeated twice.

If we were to reorganize the letters around to form another order, then we would also use factorial like in the last video. The only difference here is that since there is a repeated element, we need to divide the number of total outcomes by the repeated outcomes.




In this video, we are going to work with combinations.

Combinations: Number of orderings where order doesn’t matter.
We can use the formula: nCr\frac{n!}{r!(n-r)!} to find the combination.

For 5C3=\frac{5!}{3!\,2!}
We would have 5! on the top and 3! 2! on the bottom, which is similar to permutations with repeated elements.


Reduce if necessary

Multiply across

And we have