8.04 Combinations

In this lesson we are going to focus on problems that involve combinations. Combinations occur when the order is not important. For example:

• How many different selections can be made of two scoops of ice-cream from $8$8 flavours?
• How many different committees can be formed when choosing four people from a class of $25$25?
• How many different lottery outcomes are there of six balls chosen from $45$45 balls?
• How many different fruit smoothies can be made from a combination of banana, strawberry, watermelon and blueberries?

 

Consider the problem of selecting  $3$3 letters from the letters $A$A, $B$B, $C$C and $D$D, where we consider the selection $ABC$ABC to be the same as $BCA$BCA. In this case the order is not important. If we were counting the number of different selections of two scoops of ice-cream for a bowl, we would consider the combination chocolate and strawberry the same combination as strawberry and chocolate. 

From our formula for selecting and arranging $r$r objects from $n$n objects we need to remove all the options we counted more than once. If we were selecting $3$3 objects from $5$5 objects then we have counted each arrangement of $3$3 objects separately. To get just the number or selections, counting only $1$1 arrangement of each selection we need to divide by the number of ways to arrange the $3$3 objects which is $3!$3!. So the number of ways to select $3$3 objects from $5$5 objects is $\frac{5!}{3!\times2!}=10$5!3!×2!​=10.

In general, the number of ways to select $r$r objects from $n$n objects is $\frac{n!}{r!\left(n-r\right)!}$n!r!(n−r)!​. 

Combinations

In Mathematics, a combination is a selection of some or all of the elements of a set with no regard for order. The number of selections of $r$r objects from a set of $n$n distinct objects is given the notation $^nC_r$nCr​ or$\binom{n}{r}$(nr) or $C_r^n$Cnr​ or sometimes even $C\left(n,r\right)$C(n,r).

$^nC_r$nCr​ is often read as "$n$n choose $r$r". Since we are choosing $r$r objects from $n$n objects.

Your calculator will have options to calculate factorials and combinations. Ask your teacher if you are unsure where to find these options on your calculator.

Worked example 

Example 1

In a lottery the balls are numbered $1$1 to $45$45 and for each draw six balls are selected. How many combinations are possible?

Think: Identify $n$n and $r$r for the formula and use your calculator. We are selecting $6$6 balls from $45$45. 

Do: $n=45$n=45 and $r=6$r=6, 

$^{45}C_6$45C6​	$=$=	$\frac{45!}{6!\left(45-6\right)!}$45!6!(45−6)!​
	$=$=	$8145060$8145060

Reflect: Do these numbers reflect a real life lottery? How small are the chances of selecting a winning number?

Example 2

How many different combinations of two or three flavours of ice-cream can be selected from $8$8 possible flavours?

Think: We have an or statement, we need to add the combinations of two scoops to the number of combinations with three scoops. We are selecting $2$2 from $8$8 or $3$3 from $8$8. 

Do: 

$^8C_2+^8C_3$8C2​+8C3​	$=$=	$\frac{8!}{2!\left(6\right)!}+\frac{8!}{3!\left(5\right)!}$8!2!(6)!​+8!3!(5)!​
	$=$=	$28+56$28+56
	$=$=	$84$84

Caution: Watch for statements including 'and' or 'or'. Generally, 'and' means we need to multiply the options and 'or' means we need to add the options.

 

Practice questions

QUESTION 1

QUESTION 2

Combinations with restrictions

Let's now look at examples which have restrictions imposed to the combinations.

Worked example

Example 3

A selection of $4$4 people is to be chosen from a group of $9$9 people.  How many selections are possible if the youngest or oldest is included but not both? 

$4$4 people can be chosen from $9$9 in $^9C_4$9C4​ ways, which is $\frac{9!}{4!5!}=126$9!4!5!​=126 ways.

BUT, we have a restriction regarding the oldest or youngest.  

So what we have here is two possible situations to consider. We have the combinations where the youngest is selected plus the combinations where the oldest is selected. One of the $4$4 positions will be taken by this person, this leaves $3$3 positions to fill.  Both youngest and oldest get removed from the $9$9 people since their selections are already known.  This means that there will be $7$7 people left to choose from.  

This results in: 

$^7C_3+^7C_3=70$7C3​+7C3​=70.  

So there are $70$70 ways this selection can be made. 

 

Practice questions

QUESTION 3

QUESTION 4

QUESTION 5