Table of Contents

  1. Factorials
  2. Permutations
  3. Combinations
  4. Advanced Combination / Permutations

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!

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Common Factorials

The following list of common factorials should be memorized.

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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

permutation-combination-example3

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:

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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:

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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

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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:

  1. You multiply when you take a sequence of actions to get the desired outcome.
  2. 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:

permutation-combination-example7

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

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Case 2: Jenny is on the committee, and John is not.

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Case 3: John is on the committee, and Jenny is not.

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Total: 90 + 60 + 120 = 270


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