### Calculate probabilities using combinatorics, such as combinations and permutations

Permutations

In probability, we may be interested in the possible arrangement of a set of objects. If we are interested in the order of the arrangement of the objects, we call this arrangement a permutation.

Example: Consider flipping 2 coins. There are 4 possible permutations of this experiment (HT, HH, TH and TT).

Permutations are often shown as:

$${ _{ n }{ P }_{ r } }-permutation\quad of\quad n \quad objects\quad taken\quad r\quad at\quad a \quad time$$

The mathematical formula for a permutation can be shown as:

$${ _{ n }{ P }_{ r } }=\frac { n\left( n-1 \right) \dots \left( n-r+1 \right) \left( n-r \right) \dots \left( 3 \right) \left( 2 \right) \left( 1 \right) }{ \left( n-r \right) \dots \left( 3 \right) \left( 2 \right) \left( 1 \right) }$$

or written in another way:

$${ _{ n }{ P }_{ r } }=\frac { n! }{ \left( n-r \right) ! }$$

where $$n!$$ stands for “$$n$$ factorial” and is defined as

$$n$$ factorial – the product of all positive integers less than or equal to $$n$$

$$Example:\quad 5! = 5 \ast 4 \ast 3 \ast 2 \ast 1 = 120$$

Example

How many ways are there to select and arrange 3 letters from the letters A, B, C, D and E?

This is a permutation because we do care what order the letters are in. ABC is different than BAC, etc.

We have 5 letters and we are choosing 3 so we are looking at $${ _{ 5 }{ P }_{ 3 } }={ 5! }/{ \left( 5-3 \right) ! }={ { 5! } }/{ 2! }={ \left( 5*4*3*2*1 \right) }/{ \left( 2*1 \right) ={ 120 }/{ 2 } }=60$$

Sometimes we are interested in how many arrangements can be made from a group of objects where some of the objects are repeating.

The following formula allows us to calculate the number of arrangements of $$n$$ objects that are possible if $${n}_{1}$$, $${n}_{2}$$, etc. are alike:

$$=\frac { n! }{ { n }_{ 1 }!\ast { n }_{ 2 }!\ast \cdots { n }_{ r }! }$$

Example

How many ways are there to arrange the letters A, A, B, B, C, D and E?

This is a permutation in the form shown above. We have 7 letters of which 2 are repeating twice.

$$=\frac { n! }{ { n }_{ 1 }!\ast { n }_{ 2 }!\ast \cdots { n }_{ r }! } ={ 7! }/{ \left( 2!\ast 2! \right) }={ \left( 7\ast 6\ast 5\ast 4\ast 3\ast 2\ast 1 \right) }/{ \left( 2\ast 1\ast 2\ast 1 \right) }={ 5040 }/{ 4 }=1260$$

Combinations

If we are NOT interested in the order of the arrangement of a set of objects, we call this arrangement a combination.

Example: Consider the previous example of flipping 2 coins. There are 3 possible combinations of this experiment (2 heads, 2 tails, a tail and a head).

Combinations can be shown as:

$${ _{ n }{ C }_{ r }-n\quad objects\quad taken\quad r\quad at\quad a\quad time\quad without\quad regard\quad to\quad the\quad order\quad of\quad the\quad objects }$$

The mathematical formula for combination can be written as:

$${ _{ n }{ C }_{ r }= }\frac { n\left( n-1 \right) \dots \left( n-r+1 \right) }{ r! }$$

or written in another way:

$${ _{ n }{ C }_{ r }= }\frac { n! }{ \left( n-r \right) !\ast r! }$$

Example

How many ways are there to select a team of 5 from a group of 20 employees?

This is a combination because the order of the employees selected does not matter.

$${ _{ 50 }{ C }_{ 5 }= }\frac { 20! }{ \left( 20-5 \right) !\ast 5! } ={ 20! }/{ \left( 15!\ast 5! \right) }={ \left( 20\ast 19\ast 18\ast 17\ast 16 \right) }/{ \left( 5\ast 4\ast 3\ast 2\ast 1 \right) }=15,504$$

We can use combinations to calculate the probability of selecting certain arrangements of objects.

Example

What is the probability that we will select all hearts when selecting 5 cards from a standard 52 card deck?

The number of possible 5-card hands is 52 choose 5 or $${52!}/{(5! \ast 47!)} = 2598960$$

The number of ways to choose 5 hearts is 13 choose 5 or $${13!}/{(5!*8!)} = 1287$$

So, the probability of selecting all hearts is $${1287}/{2598960} = .000495 \quad or \quad .0495\%$$

Learning Outcome

Topic 1.f: General Probability – Calculate probabilities using combinatorics, such as combinations and permutations.