Arrangements & Factorials (Cambridge (CIE) A Level Maths: Probability & Statistics 1): Revision Note

Arrangements

How many ways can n different objects be arranged?

• When considering how many ways you can arrange a number of different objects in a row it’s a good idea to think of how many of the objects can go in the first position, how many can go in the second and so on

• For $n=2$ there are two options for the first position and then there will only be one object left to go in the second position so

  • To arrange the letters A and B we have

    • AB and BA

  • For $n=3$  there are three options for the first position and then there will be two objects for the second position and one left to go in the third position so

    • To arrange the letters A, B and C we have

      • ABC, ACB, BAC, BCA, CAB and CBA

    • For n objects there are $n$ options for the first position, $n-1$ options for the second position and so on until there is only one object left to go in final position

    • The number of ways of arranging different objects is $n\times (n-1)\times (n-2)\times ...\times 2\times 1$ 

What happens if the objects are not all different?

• Consider arranging two identical objects, although there are still two different ways you could place the objects down next to each other, the arrangements would look exactly the same

• To arrange the letters A­₁ and A₂ we have

  • A­₁ A₂ and A₂ A­­_­1

  • These are exactly the same, so there is only one way to arrange the letters A and A

• To arrange the letters A₁, A₂ and C we have

  • A­₁ A₂ C, A­₂ A­₁ C, A­₁ ­­­C A­₂, , A­₂ C A₁, C A­₁ ­­­A­₂, C A­₂ A­₁

  • Although the two letter As were placed separately, they are identical and so each pattern has been repeated twice

  • There are 6 ways to arrange the letters A, A and C, but with some duplicates

  • There are $\frac{6}{2}=3$ different ways to arrange the letters A, A and C   

• If there are two identical objects within a group of objects to be arranged, the number of ways of arranging different objects should be divided by 2

• Consider arranging three identical objects, although there are still six different ways you could place the objects down next to each other, the arrangements would look exactly the same

• To arrange the letters A₁, A₂ and A₃

  • A­₁  A₂ A­₃ , A₁  A₃  A­₂  , A­₂  A₁ A­₃ , A₂ A₃ A­₁ , A­₃ A₁ A­₂  ,  A₃ A₂ A­₁

  • However, if these were all A, we would have AAA repeated six times

• To find the number of arrangements of the letters A, A, A and C we would have to consider the number of ways of arranging four letters if they were all different and then divide by the number of ways AAA is repeated

  • Four different letters could be arranged $4\times 3\times 2\times 1$ = 24 ways

  • AAA would be repeated six times so we would need to divide by 6

  • There are four different ways to arrange the letters A, A, A and C 

• If there are three identical objects within a group of n objects to be arranged, the number of ways of arranging $n$ different objects should be divided by 6

• If there are r  identical objects within a group of n objects to be arranged, the number of ways of arranging $n$ different objects should be divided by the number of ways of arranging r different objects

  • If there are r  identical objects within a group of n objects to be arranged, the number of ways of arranging n the objects is $n\times (n-1)\times (n-2)\times ...\times 2\times 1$  divided by  $r\times \left(r-1\right)\times \left(r-2\right)\times ...\times 2\times 1$

Factorials

What are factorials?

• Factorials are a type of mathematical operation (just like +, -, ×, ÷)

• The symbol for factorial is !

  • So to take a factorial of any non-negative integer, $n$ , it will be written $n$!  And pronounced ‘ $n$ factorial’

• The factorial function for any integer, $n$, is $n!=n\times (n-1)\times (n-2)\times ...\times 2\times 1$

  • For example, 5 factorial is 5! = 5 × 4 × 3 × 2 × 1

• The factorial of a negative number is not defined

  • You cannot arrange a negative number of items

• 0! = 1

  • There are no positive integers less than zero, so zero items can only be arranged once

• Most normal calculators cannot handle numbers greater than about 70!, experiment with yours to see the greatest value of $x$ such that your calculator can handle $x!$

How are factorials and arrangements linked?

• The number of arrangements of $n$ different objects is $n!$

  • Where $n!=n\times (n-1)\times (n-2)\times ...\times 2\times 1$

• The number of different arrangements of $n$objects with one object repeated $r$ times and the others all different is $\frac{n!}{r!}$

• The number of different arrangements of $n$objects with one object repeated $r$ times, another object repeated $s$times and the other objects all different is $\frac{n!}{r!s!}$

What are the key properties of using factorials?

• Some important relationships to be aware of are:

  • $n!=n\times (n-1)!$

    • Therefore

$\frac{n!}{(n-1)!}=n$

• $n!=n\times (n-1)\times (n-2)!$

  • Therefore

$\frac{n!}{(n-2)!}=n\times (n-1)$

• Expressions with factorials in can be simplified by considering which values cancel out in the fraction

  • Dividing a large factorial by a smaller one allows many values to cancel out

$\frac{7!}{4!}=\frac{7\times 6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1}=7\times 6\times 5$