Combination and Permutation of Objects

1. Permutation

A permutation is an arrangement of objects where the order is important.

Formula

nPr = (n!) / ((n-r)!)

Where:

• n = total number of objects

• r = number of objects selected

• n! = factorial of n

Example 1.

Three students (Ana, Ben, and Carl) compete for:

• 1st place

• 2nd place

• 3rd place

Possible arrangements:

1. Ana, Ben, Carl

2. Ana, Carl, Ben

3. Ben, Ana, Carl

4. Ben, Carl, Ana

5. Carl, Ana, Ben

6. Carl, Ben, Ana

There are 6 permutations because changing the order creates a different arrangement.

nPr = (n!) / ((n-r)!)

3P3 = (3!) / ((3-3)!) 

3P3 = (3x2x1) / (0!) 

3P3 = (6) / (1) 

3P3 = 6

Using Calculator:

3P3 = 6

2. Combination

A combination is a selection of objects where the order is not important.

Formula

nCr = (n!) / (r! (n-r)!)

Example 2.

Choose 2 students from:

• Ana

• Ben

• Carl

Possible selections:

1. Ana and Ben

2. Ana and Carl

3. Ben and Carl

There are 3 combinations.

Notice that:

• Ana and Ben

• Ben and Ana

represent the same group, so they are counted only once.

nCr = (n!) / (r! (n-r)!)  

3C2 = (3!) / (2! (3-2)!)  

3C2 = (3x2x1) / (2! 1!) 

3C2 = (6) / (2!) 

3C2 = (6) / (2x1) 

3C2 = (6) / (2) 

3C2 = 3

Using Calculator:

3C2 = 3

Calculating the Number of Combinations and Permutations of n Objects Taken k at a Time

Example 1.

Find the number of permutations of 5 objects taken 3 at a time.

nPr = (n!) / ((n-r)!)

5P3 = (5!) / ((5-3)!) 

5P3 = (5!) / (2!) 

5P3 = (5x4x3x2x1) / (2x1) 

5P3 = (120) / (2) 

5P3 = 60 permutations

Example 2.

From the same 8 students, how many ways can a President, Vice President, and Secretary be selected?

nPr = (n!) / ((n-r)!)

8P3 = (8!) / ((8-3)!) 

8P3 = (8!) / (5!) 

8P3 = (8x7x6x5x4x3x2x1) / (5x4x3x2x1) 

8P3 = (40320) / 120 

8P3 = 336 arrangements

Example 3.

Find the number of combinations of 5 objects taken 3 at a time.

nCr = (n!) / (r! (n-r)!)

5C3 = (5!) / (3! (5-3)!) 

5C3 = (5!) / (3! 2!) 

5C3 = (5x4x3x2x1) / ((3x2x1) (2x1)) 

5C3 = (120) / ((6)(2)) 

5C3 = (120) / (12) 

5C3 = 10 combinations

Example 4.

A class has 8 students. How many ways can 3 students be chosen for a committee?

nCr = (n!) / (r! (n-r)!)

8C3 = (8!) / (3! (8-3)!) 

8C3 = (8!) / (3! 5!) 

8C3 = (8x7x6x5x4x3x2x1) / ((3x2x1) (5x4x3x2x1)) 

8C3 = 40320 / ((6) (120)) 

8C3 = 40320 / 720 

8C3 = 56 committees