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13.10 Probabilities with combinations

Lesson

The word combination can be used in both of these examples:

  • The combination to the lock is $358$358
  • I made a salad using a combination of lettuce, tomato and onion.

What's the different between these two situations? It all comes down to whether order matters.

Obviously for the salad, it doesn't matter which order the ingredients go in - but it is a very different story for the lock.  Any other order like $853$853 or $385$385 for example will not open the lock. 

This is the difference between our permutations and combinations.  Permutations are like the lock, the order matters and we treat the $853$853 as different to $358$358.  Combinations are where the order doesn't matter so we treat the lettuce, tomato and onion salad as the same as a salad with onion, lettuce and tomato. 

As we have already looked at permutations and a way to calculate total possible arrangements, we can consider permutations as an ordered version of a combination. 

Worked example

Question 1

Suppose I have a red dot, blue dot and green dot.

A) How many ordered arrangements can I make if I select exactly two dots?

Think: For an arrangement, order matters, so this is a permutation.

Do:

 

B) How many different combinations of two different colored dots are there?

Think: For a combination order does not matter, so red-green is the same as green-red.

Do:

 

Reflect: If order mattered, there were $6$6 possibilities, if order did not matter then there were only $3$3 possibilities. This is because for every combination of $2$2 dots, there were $2$2 possible orders or arrangements.

 

This idea of counting how many possible different combinations there are is part of a mathematical topic called Counting Techniques.  For small sets like the one above, it's quite simple to write down all the choices and add them up.  But for larger sets, like imagine we had $8$8 colors and wanted to see how many different combinations of $5$5 colors there are - we need a more efficient method of counting.  

 

Formula for combinations

The lotto (or lottery) can be an example of a combination.  It is a combination if it doesn't matter what order the balls are drawn, or what order you picked your numbers as long as at the end of the draw, the numbers on your ticket match the numbers that were drawn. 

Let's work through an example of a very simple lotto game.  There are $10$10 balls and we can win $\$400$$400 if the $3$3 numbers we choose are drawn.

How many arrangements? How many possible ways can we draw $3$3 balls from $10$10?

$10$10 $9$9 $8$8

We have $10$10 possibilities for the first ball, then $9$9 for the second and then finally $8$8 for the third. 

We could have also used the formula for permutations:

$\nPr{10}{3}$10P3 $=$= $\frac{10!}{\left(10-3\right)!}$10!(103)!
  $=$= $\frac{10!}{7!}$10!7!
  $=$= $10\times9\times8$10×9×8
  $=$= $720$720

The answer to this is $\nPr{10}{3}=720$10P3=720.

So there are $720$720 arrangements.  But in this list would be options like $5,6,7$5,6,7 and $5,7,6$5,7,6 and $6,7,5$6,7,5 and $6,5,7$6,5,7 and $7,5,6$7,5,6 and $7,6,5$7,6,5..... all of which are actually the same with regards to our lotto game.  

So what we want to do is remove all these possible double ups.

As we are choosing $3$3 balls, there are $3!=3\times2\times1=6$3!=3×2×1=6 times as many arrangements than we need.  

So we take our permutations formula  $\frac{n!}{(n-r)!}$n!(nr)! and divide through by $r!$r! (to remove the double ups).

Combinations formula

Where $n$n and $r$r are positive integers, $n\ge r$nr, and $n$n is the total number of elements in the set and $r$r is the number to be chosen:

$_nC_r=\frac{n!}{r!(n-r!)}$nCr=n!r!(nr!)

 

So for our lotto game, 

$C\left(10,3\right)=\frac{10!}{3!7!}=\frac{10\times9\times8\times7!}{3!7!}=\frac{10\times9\times8}{3\times2\times1}=\frac{720}{6}=120$C(10,3)=10!3!7!=10×9×8×7!3!7!=10×9×83×2×1=7206=120 different combinations. 

So we have a $1$1 in $120$120 chance of winning.   

 

Notation for combinations

There are four possible notations we can see in different books for choosing $r$r objects from a set of $n$n or the formula $\frac{n!}{r!\left(n-r\right)!}$n!r!(nr)!. For all of them we say "$n$n choose $r$r".

$_nC_r=\frac{n!}{r!(n-r!)}$nCr=n!r!(nr!)

$^nC_r=\frac{n!}{r!(n-r!)}$nCr=n!r!(nr!)

$\binom{n}{r}=\frac{n!}{r!(n-r!)}$(nr)=n!r!(nr!)

$C\left(n,r\right)=\frac{n!}{r!(n-r!)}$C(n,r)=n!r!(nr!)

 

Properties of combinations

Using the definition of factorials we can show a few interesting properties of combinations

  1. We can show that there are the same number of ways to choose $r$r items from $n$n as there are to choose $\left(n-r\right)$(nr) items from $n$n.
$C\left(n,r\right)$C(n,r) $=$= $\frac{n!}{r!\times\left(n-r\right)!}$n!r!×(nr)!
  $=$= $\frac{n!}{\left(n-r\right)!\times r!}$n!(nr)!×r!
  $=$= $C\left(n,n-r\right)$C(n,nr)

This means for example that $\nCr{12}{4}=\nCr{12}{8}$12C4=12C8 and $nCr(7,1)=nCr(7,6)$nCr(7,1)=nCr(7,6). There is a nice symmetry with this expression. 

 

  1. We can show that there is only $1$1 way to choose $n$n items from $n$n items or $0$0 items from $n$n items.

It is important to note that by definition $0!=1$0!=1 and $1!=1$1!=1, we have by definition that

$C\left(n,0\right)$C(n,0) $=$= $\frac{n!}{0!\times\left(n-0\right)!}$n!0!×(n0)!
  $=$= $\frac{n!}{1\times n!}$n!1×n!
  $=$= $1$1

For example, $C\left(5,0\right)=1$C(5,0)=1 and $C\left(12,12\right)=1$C(12,12)=1.

 

  1. Finally we can form a relationship between $C(n,r)$C(n,r) and the previous $P(n,r)$P(n,r) as follows:
$C\left(n,r\right)$C(n,r) $=$= $\frac{n!}{r!\times\left(n-r\right)!}$n!r!×(nr)!
  $=$= $\frac{n\times\left(n-1\right)\times\left(n-2\right)\times...\times\left(n-r+1\right)\times\left(n-r\right)!}{r!\times\left(n-r\right)!}$n×(n1)×(n2)×...×(nr+1)×(nr)!r!×(nr)!
  $=$= $\frac{n\times\left(n-1\right)\times\left(n-2\right)\times...\times\left(n-r+1\right)}{r!}$n×(n1)×(n2)×...×(nr+1)r!
  $=$= $\frac{P\left(n,r\right)}{r!}$P(n,r)r!

So $C\left(n,r\right)=\frac{P\left(n,r\right)}{r!}$C(n,r)=P(n,r)r! and this is another important mathematical result in the study of probability. 

 

Practice questions

QUESTION 2

Evaluate $\nCr{7}{2}$7C2.

QUESTION 3

A boss wants to select one group of $4$4 people from his $28$28 staff.

How many different groups are possible?

 

Let's look at some more involved questions. This often includes questions where we impose some restrictions on the combinations.  

Worked example

Question 4

A selection of $4$4 people are 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? 

Think - Strategy 1:

$4$4 people can be chosen from $9$9 in $\nCr{9}{4}$9C4 ways, which is $\frac{9!}{4!5!}=126$9!4!5!=126 ways. However, we have a restriction regarding the oldest or youngest.  

So what we have here is two possible situations to consider. In each case, one person must definitely fill one of the $4$4 positions, 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.  

Do - Strategy 1:

Case including youngest + Case including oldest $=$= $\nCr{7}{3}+\nCr{7}{3}$7C3+7C3
  $=$= $2\nCr{7}{3}$27C3
  $=$= $2\times\frac{7!}{3!4!}$2×7!3!4!
  $=$= $2\times\frac{210}{6}$2×2106
  $=$= $2\times35$2×35
  $=$= $70$70

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

Think - Strategy 2:

Sometimes it helps to have a method that keeps track of the whole situation. Let's also use another notation to get used to it:

  • The first thing to occur, is that we choose the oldest, we do that in $\binom{1}{1}$(11) ways.
  • The second thing is then to ignore the youngest, we do that in $\binom{1}{0}$(10) ways
  • The third thing is then to choose the remaining $3$3, and we do that in $\binom{7}{3}$(73) ways. 

When we write these as a multiplication, we get 

$\binom{1}{1}\binom{1}{0}\binom{7}{3}$(11)(10)(73)

Look at the top row here (the $n$n positions).... $1+1+7=9$1+1+7=9 which is the total number of people we had to choose from

Look at the bottom row here (the $r$r positions).... $1+0+3=4$1+0+3=4 which is the total number of people we were choosing.  

This method is great to keep track of everyone.  We don't want to leave anyone out! 

Do - Strategy 2:

$\binom{1}{1}\binom{1}{0}\binom{7}{3}+\binom{1}{1}\binom{1}{0}\binom{7}{3}=2\times\binom{1}{1}\binom{1}{0}\binom{7}{3}=1\times1\times70=70$(11)(10)(73)+(11)(10)(73)=2×(11)(10)(73)=1×1×70=70

 

Practice questions

QUESTION 5

A team of $3$3 is to be chosen at random from a group of $5$5 girls and $6$6 boys.

In how many ways can the team be chosen if:

  1. there are no restrictions?

  2. there must be more boys than girls?

QUESTION 6

$5$5 boys and $6$6 girls are part of the debate team. $4$4 of them must represent the school at the upcoming debate tournament. In how many ways can the team of $4$4 be formed if:

  1. it must contain $2$2 boys and $2$2 girls?

  2. it must contain at least $1$1 boy and $1$1 girl?

 

Probabilities with combinations

Recall that to find the probability of something occurring we consider the two values:  the number of ways the thing you want can happen and the total number of ways it could happen.

Let's consider calculating some probabilities of events with sample spaces that include combinations. We still approach the calculations of the combinations the same, but then we have the added step of calculating the probabilities. 

Worked example

Question 7

A magazine editor is deciding which of $5$5 articles to print as the cover story on the front page.  

a. If only $2$2 stories can be chosen, how many different selections are possible?

Think: First, we confirm that it is a combination. Since the order is important, we know now we need to use the combination formula.  We check the question looking for restrictions, but there aren't any.  Therefore, we will use the formula

$C(n,r)=\frac{n!}{r!(n-r)!}$C(n,r)=n!r!(nr)!

Where $n$n is the total number of articles $n=5$n=5 and $r$r is the number we are choosing, $r=2$r=2

Do:

$C(5,2)$C(5,2) $=$= $\frac{5!}{2!(5-2)!}$5!2!(52)!
  $=$= $\frac{5!}{2!(3)!}$5!2!(3)!
  $=$= $\frac{5\times4\times3!}{2!3!}$5×4×3!2!3!
  $=$= $\frac{5\times4}{2!}$5×42!
  $=$= $10$10
  $=$=  

So there are $10$10 combinations.

 

b. If you were the writers of one of the stories, what is the probability that your article is randomly chosen for the front page?

Think: If your article is chosen, then it is definitely one of the two. So there are now 4C1 possible combinations of your article (and one other) being on the front cover.  

Total combinations of your article appearing = $C(4,1)=4$C(4,1)=4

Do: Thus the probability becomes, 

$\text{Probability of Your Article}=\frac{\text{What you want}}{\text{Total possible}}\times100%$Probability of Your Article=What you wantTotal possible×100%

$\text{Probability of Your Article}=\frac{4}{10}\times100%=40%$Probability of Your Article=410×100%=40%

 

Practice questions

QUESTION 8

A box contains 6 pens of different colors: red, green, blue, yellow, black and white. Two pens are drawn at random without replacement.

  1. How many possible selections are there?

  2. What is the probability of drawing the green and black pens?

QUESTION 9

A menu has three entrées ($E_1,E_2,E_3$E1,E2,E3), four mains ($M_1,M_2,M_3,M_4$M1,M2,M3,M4) and two desserts ($D_1,D_2$D1,D2). A meal is made up of one of each.

  1. How many different meals are possible?

  2. What is the probability of selecting $E_1$E1 , $M_3$M3 and $D_2$D2?

  3. How many different meals are possible given that $E_1$E1 is the entrée?

QUESTION 10

$5$5 people are to be selected from a larger group of $10$10 candidates. If Amelia is among the candidates, what is the probability that she will be among those selected?

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