Table of Contents
Factorials
A factorial is distinguished by a number followed by an exclamation mark (!). The factorial of a variable n is the product of all of the whole numbers from 1 to n.
Example
Evaluate 6!
Common Factorials
The following list of common factorials should be memorized.
Permutations
Loosely speaking, permutations are used to display the total number of outcomes given a certain fact. In permutations, order matters. Permutations can be calculated using the formula
Combinations
Combinations are very similar to permutations. The only difference is that in combinations, order does not matter. The formula to calculate the number of possible combinations is:
Understanding the specific differences between permutations and combinations is not important. It’s only important to know which equation to apply depending on the problem statements.
Example
A thief is trying to crack an atm machine. The keypad consists of 10 numbers (0-9). If no numbers can be repeated and the atm password is 4 characters long, how many unique combinations exist?
First determine whether this is a permutation or combination question. Since changing the order of numbers gives us a new combination, order matters. This is a permutation question.
n = 10, because the thief is choosing from a total of 10 possible numbers.
k = 4, because each atm password is 4 characters long.
Apply the formula:
There are 5040 possible combinations
Example
A manager is selecting 3 people out of a pool of 7 volunteers to form a new safety committee. How many different committees can be formed?
Determine whether this is a permutation or combination problem. Since changing the order of the members on a team would still give us the same team, order does not matter. This is a combination problem.
n = 7, because there are a total of 7 volunteers
k = 3, because we are trying to form a committee of 3 members
There are 35 different committees that can be formed
Advanced Combination / Permutations
The most difficult combination and permutation problems involve constraints. In these advanced difficulty problems, a best practice approach is to list the winning scenarios. There are two rules to remember:
- You multiply when you take a sequence of actions to get the desired outcome.
- You add when you break the outcome into different cases. You add the cases to get the total.
Example
From a group of 7 men and 5 women, how many different committees consisting of 4 men and 2 women can be formed?
Determine whether this is a permutation or combination problem. Since changing the order of the members on a committee would still give us the same committee, order does not matter. This is a combination problem.
This problem follows a sequence of actions: first choose the men, and then choose the women. Apply this sequence into the combinations formula:
There are 350 committees that can be formed.
Example
From a group of 7 men and 5 women, how many different committees consisting of 4 men and 2 women can be formed if Jenny and John refuse to be on the committee together?
Determine whether this is a permutation or combination problem. Since changing the order of the members on a committee would still give us the same committee, order does not matter. This is a combination problem.
Deconstruct the problem into different cases and sum together the possible combinations of all the different cases to find the total number of committees possible.
Case 1: Both Jenny and John are not on the committee.
Sequence of actions: choose 4 men from the remaining 6 men; choose 2 women from the remaining 4 women
Case 2: Jenny is on the committee, and John is not.
Case 3: John is on the committee, and Jenny is not.
Total: 90 + 60 + 120 = 270
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