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Permutations without formulas or tears
Robert Yen, Ambarvale High School
In how many ways can 5 boys and 4 girls sit in a row if Katie and Christine want to sit together?
Do these kinds of problems intimidate and confuse you? Permutations and combinations was the one topic that caused me anxiety when I was studying for my HSC. It made little sense to me, and like many students with a probability phobia, I could never be sure if my final answer was correct! And judging by the number of calls to the HSC Advice Line on this very subject, many students today still experience a lack of confidence when using counting techniques to solve probability problems.
To overcome my limitations, I researched and studied this topic thoroughly the first time I taught a 3 Unit class. I knew there had to be a more straightforward, intuitive approach to teaching this topic, hopefully one free of jargon and formula. I'm happy to report that there is, and this became most self-evident some years later on a day when half of my Year 11 class were away on an excursion. I decided to give the remaining half a quick preview of their next 'difficult' topic, but surprised them and myself when I covered most of the permutations and combinations theory in that hour, and without mentioning or .
When I related this story to my head teacher afterwards, he agreed that yes, permutations is best taught using a 'lists-and-boxes' problem-solving approach, starting from basic principles and relying less upon fancy formulas. What follows are extracts from my teaching notes, which have been refined several times, with the hope that they may help you alleviate some of your students' fears (and perhaps your own) next time you teach this topic.
Introduction
There are five horses in a race - A, B, C, D and E. Suppose you have to bet on which two horses come first and second, in the correct order. From the five horses, how many different bets (1st - 2nd pairings) are possible?
AB BA CA DA EA
AC BC CB DB EB
AD BD CD DC EC
AE BE CE DE ED
Number of arrangements = 20. These arrangements are called permutations.
Now suppose instead that you need to choose the two horses that come first or second, but you don't have to state which one comes first and which once comes second. In other words, the order is not important. Now how many selections are possible?
AB BC CD DE AC
BD CE AD BE AE
Number of selections = 10. These selections are called combinations.
Notice how there are only half as many combinations as permutations because order is not important with combinations (e.g., AB is the same as BA).
The multiplication principle for counting arrangements
1.From this list of given names and surnames, how many different arrangements of given name - surname pairs are possible?
Given name SurnameAlex GarrettBrionne HinzeCathy IbsenDarren JohnsonErin KingFiona
Using a tree diagram, but not completing it ...
For each given name, there are 5 possible surnames.
Total possible arrangements is 6 x 5 = 30.
2. From this menu, calculate how many different 3-course dinners are possible.
Entrée Main course DessertPumpkin soup
Garlic prawns Pavlova
Calamari rings
Steak Diane Black Forest cake
Lasagne Roast lamb Chocolate mousse
Chicken Dijon
Mangoes and ice cream
Grilled perch
Total possible dinner arrangements is 3 x 5 x 4 = 60 (and I've tried every one!!).
The basic counting principle is:
1.If A can be arranged in m different ways and B can be arranged in n different ways, then the number of possible arrangements of A and B together is m x n.
2.More generally, if
A can be arranged in a different ways,
B can be arranged in b different ways,
C can be arranged in c different ways, and so on;
then the total number of possible arrangements of ABC ... together is a x b x c ....
Permutations
1. A girls' school is electing a captain and vice-captain. There are four candidates: Amy, Betty, Caroline and Deane. How many possible arrangements of captain/vice-captain are there?
Instead of using a tree diagram to count the possibilities, we can draw boxes for the two positions.
There are 4 ways of choosing the captain, but once the captain is determined, there are only 3 ways of choosing the vice-captain.Total possible arrangements = 4 x 3 = 12.
These arrangements are called permutations. The number of permutations possible has the notation where n is the number of items or people available, and r is the number of places available. In the above example there were 4 candidates and 2 places.
.
Note is found by multiplying backward from n, r times.
2.8 people are to be photographed, but there are only 5 seats available. In how many ways can 8 people be seated in 5 seats?
8 people, 5 places.
arrangements.
Permutations with restrictions
In mathematics, '!' (read 'factorial') means a special product.
For example, 5! = 5 x 4 x 3 x 2 x 1 ('5 factorial')
x! means multiplying backward from x down to 1. There is also an x! key on your calculator.
This factorial notation is good for abbreviating numerical expressions in permutation problems.
1.In how many different ways can 5 boys and 4 girls sit together in a row?
2.What if boys and girls must alternate?
3.What if boys and girls sit in separate groups?
4.What if Katie and Christine want to sit together (they always do)?
We can solve this problem by drawing boxes to represent seats or positions. Boxes are like an advanced representation of the levels (branches) of a tree diagram.
1.9 seats, 9 places. arrangements.
9 8 7 6 5 4 3 2 1
2.
B G B G B G B G B
5 4 4 3 3 2 2 1 1
Boys: 5 seats, 5 places, Girls: 4 seats, 4 places.
5! x 4! = 2880 arrangements.
3.
B B B B B G G G G
or
G G G G B B B B B
2 x 5! 4! = 2 x 2880 = 5760 arrangements.
4. Let both girls be considered as one person/seat.
Now there are only '8 people' to arrange on 8 seats. 8! = 40 320.
But for the '2-girls seat' GG, the two girls can sit in 2 different ways: G1G2 or G2G1 (Katie and Christine can swap seats). So for each of the 8! arrangements, there are 2 possibilities for GG.
2 x 8! = 80 640 arrangements.
Permutations with special conditions
(a) Arrangements around a circle
In how many ways can 6 people be arranged around a table, if the order around the table is all that matters?
This is different to 6 people in a line, because the 6 people in a circle can all move one (or more) places to the left, without affecting the order of seating. In other words, A B C D E F, B C D E F A, C D E F A B, etc., are ALL the same circular arrangement because they have the same order.
If the seats were in a line, there would be 6! = 720 possible arrangements. However, for each of these arrangements, you can move every person left one seat and the order would still be retained. In fact, you can continue doing this until you get back to where you started - that would be 6 shifts. The answer 6! is 6 times too high, so we must divide by 6.
Total possible arrangements =
Generally, n items in a circle (e.g. seats, keys on a key ring) can be arranged in (n - 1)! ways.
Alternative method: Another way of thinking of the above example is to say that it doesn't matter where the first person (X) sits, since only the order around him/her is important. Total possible arrangements = 5! = 120.
(b) Permutations of n items, some alike
1.How many 7-letter permutations are possible from the letters of the word LOLLIES?
Solution
The answer is less than 7!, because L is a repeated letter, so some of the arrangements will be the same, for example, E I L1 L2 L3 O S, E I L1 L2 L3 O S, E I L1 L2 L3 O S, etc., are the same.
For each arrangement of 7 letters, the 3 L'sE I L L L O S
can be arranged in 3! ways, so the 7! needs to be divided by 3!.
No. of permutations =
From n items, of which p are alike of one kind, q are alike of another kind, and so on,
the number of possible permutations is
2. (1989 HSC) Let each different arrangement of all the letters of DELETED be called a word.
(a) How many words are possible?
(b) In how many of these words will the D's be separated?
Solution
(a) DELETED has 2 D's, 3 E's. Permutations
(b) To count how many words have separated D's, we can count how many words have D's together and subtract from (a). Let DD be counted as one letter, giving 6 letters.
Permutations
No. of words with separated D's = 420 - 120 = 300.
Combinations
Three students need to be selected from a group of 8 students to represent the school at a conference.
How many combinations of 3 representatives are possible?
8 students, 3 places. There are permutations, but this isn't the answer to the problem. If A, B, C, D, ... represent the 8 students, then the permutations ABC, ACB, BAC, BCA, CAB, CBA are the same selection. These 3 students can change places within the group but it's still the same group. Because of this, we must divide our permutations answer by 3! = 6.
No. of combinations
Combinations are a special type of permutation. The number of combinations is smaller because the order of arrangements is not important, e.g. ABC = ACB = BAC = .... , so we divide by r! where r is the number of places available.
Combinations have the formula and notation where n is the number of items available and r is the number of places available.
So for selecting 3 students from 8 students to go to a conference:
Permutations vs combinations
Permutations are arrangements of items, while combinations are selections of items. When you arrange something, you organize them in a particular order or place, but when you select something, you simply pick out or choose them, without considering their order or place. When selecting from a group of items, the order of the items selected is not important. is sometimes called 'n choose r', for example, is
sometimes read '8 choose 3'. Sometimes, is also written as , for example, can
also be written .
Permutations are ordered arrangements: there are more of them because ABC, ACB, BAC, etc. are considered different. For example, a 'trifecta' bet on a horse race (first 3 horses in correct order) is a permutation. Combinations are unordered selections: there are fewer of them because ABC, ACB, BAC, etc. are considered the same. For example, a Lotto bet (6 numbers to come up in any order) is a combination.
Formulas for and
Formulas for and can be written using x! notation.
For example:
Generally,
for example:
Generally:
Most calculators also have and keys, so permutations and combinations can be calculated in three different ways: by multiplying backward, by using the n! formulas, or by using the calculator keys.
Investigations
1. Why is nPn = n!?
2. Why is nCn = 1?
3. Why is nCr = nCn-r? (symmetrical property)
4. Why is nC1 = n?
5. Why is nP1 = n?
Combinations with restrictions
1. (1996 HSC) A committee of 3 men and 4 women is to be formed from a group of 8 men and 6 women. In how many ways can this be done?
Solution
No. of ways 3 men can be chosen from a group of 8 men = .
No. of ways 4 women can be chosen from a group of 6 women = .
No. of possible committees
2.Three letters are randomly chosen from the word TUESDAY. How many possible selections are there? How many of these selections have:
(a) exactly one vowel?
(b) exactly 2 vowels?
(c) at least 2 vowels?
Solution
No. of selections
(a) 'Exactly one vowel 'means 1 vowel and 2 consonants. Drawing dashes to represent these:
where V = vowel, and = not vowel.
TUESDAY has 3 vowels and 4 consonants.
No. of selections =
Note: For convenience, we'll use boxes for permutations and dashes for combinations.
(b) 'Exactly 2 vowels':
No. of selection
(c) 'At least 2 vowels' means 2 or 3 vowels.
No. of selections of 3 vowels Total selections = 12 + 1 = 13.
Binomial probability
With binomial probability, we are concerned with repeated probability trials in which there are only two possible outcomes: a success with probability p, and a failure with probability q, where p + q = 1.
These are also called Bernoulli trials after the Swiss mathematician James Bernoulli (1654 -1705). Some examples of outcomes in Bernoulli trials are HEAD or TAIL, WIN or LOSE, TRUE or FALSE, BOY or GIRL, DEFECTIVE or NON-DEFECTIVE.
The probability of hitting a target is . Garrett takes 5 shots at the target.
Calculate the probability that he scores:
(a) no hits
(b) only 1 hit
(c) 2 hits
(d) 3 hits
(e) 4 hits
(f) 5 hits.
Solution
Drawing a tree diagram but not completing it ...
Let X represent the number of hits out of 5.
(a) x x x x x
No. of different ways of Getting 1 correct from 5.
x x x x
No. of different ways of Getting 2
x x x
correct from 5.
(d) x x
(e) x
(f)
The general pattern is where n is the total number of trials and r is the number of successes. Note that the sum of the probabilities equals 1.
If p = P(success) and q = P(failure) = 1 - p, and the Bernoulli trial is repeated n times, then the probability of r successes is
A SYSTEMATIC APPROACH TO PERMUTATIONS AND COMBINATIONS
Paul Larkin
Introduction
The following approach to solving permutation and combination problems results from the difficulty that my students and I encountered when I first taught it. Deciding how to solve a particular problem involves asking a set of questions. The responses to these questions allow the problem to be placed in a category that indicates the method of solution. This paper contains these questions and categories, with worked examples and some challenging problems that do not fit into any particular category.
Questions and categories
The questions students need to ask are:
1. Once an item has been chosen, is it replaced?
2. Are restrictions placed on how the items are chosen?
3. Is the order in which items are chosen important?
4. If order is important, are the items being arranged in a line or a circle?
For items arranged in a line without restriction, there are these questions:
5. Are all the items being arranged?
6. If all the items are being arranged, is each item unique?
Question 1 reminds us that permutations and combinations are used when items are chosen without replacement. Question 2 breaks problems into two categories, 'without restriction' and 'with restriction'. Questions 3 and 4 give three subcategories: line permutation, circle permutation and combination. Unrestricted line permutations are of three types: all items are arranged and are unique; all items are arranged and some are identical; only some items are arranged. These categories are shown in Figure 1.
FIGURE 1
Solutions - without restriction
FIGURE 2
Figure 2 shows the types of questions that fall within the category 'without restriction' and the general method of solution. Here are some specific examples:
Line permutation
1. All items, each unique
Q. In how many ways can seven students be arranged in a line?
Order important - line permutation, without restriction.
All 7 items arranged and unique.
No. of arrangements = 7! = 5040.
2. All items, some identical
Q. In how many ways can the letters of the word WOLLONGONG be arranged?
Order important - line permutation, without restriction.
All 10 items arranged.
Some items identical: 3 Os, 2 Ls 2 Ns, 2 Gs.
No. of arrangements
3. Some items
Q. How many different arrangements are there using 3 letters of the word SUNDAY?
Order important - line permutation, without restriction.
No. of items = 6.
No. of items chosen = 3.
No. of arrangements
Circle permutation
Q. In how many ways can six people be arranged in a circle?
Order important - circle permutation, without restriction.
No. of items = 6.
No. of arrangements = (6 - 1)! = 120.
Combinations
Q. How many committees of 4 students can be chosen from 8 students?
Order unimportant - combination.
No. of students = 8.
No. of students chosen = 4.
No. of committees
Solutions - with restriction
Note: These strategies are not rules or formulas. Carefully adapt each strategy to the problem you are solving.
FIGURE 3.
Figure 3 shows the types of problems that occur with restriction and the general approach used to solve each type. Here are three important things to note:
(a) The presence of a restriction is usually but not always indicated by the word 'if'.
(b) A restriction breaks the items into two groups:
a group of restricted items and a group of unrestricted items (see Q1), or two restricted groups of items (see Q2).
(c) With line permutations there are two kinds of restrictions:
an item's position is restricted in relation to other items (see Q1, Q2); a location is restricted to only being occupied by some items (see Q3).
Line permutation
Q1. How many ways can 4 boys and 3 girls be arranged in a line so that all 4 boys are together?
Order important - line permutation, with restriction.
Restricted items - 4 boys, unrestricted items - 3 girls.
Restriction - 4 boys together:
B B B B G G G G B B B B G G G G B B B B G G G G B B B B,
that is, 4 patterns.
Arr. of 4 boys = 4! = 24.
Arr. of 3 girls = 3! = 6.
Total no. of arr. = 4 x 24 x 6 = 576.
Q2. How many lines of 4 boys and 3 girls are there if boys and girls alternate?
Order important - line permutation, with restriction.
Restricted items - 4 boys, 3 girls.
Restriction - boys and girls alternate: B G B G B G B,
that is, 1 pattern.
Arr. of 4 boys = 4! = 24.
Arr. of 3 girls = 3! = 6.
Total no. of arr. = 1 x 24 x 6 = 144.
Q3. In how many ways can 2 adults and 3 children be arranged in a line if the adults are at either end of the line?
Order important - line permutation, with restriction.
Restricted items - 2 adults, unrestricted items - 3 children.
Restriction - 2 adults at either end of line: A - - - A.
Arr. of 2 adults = 2! = 2.
Arr. of 3 children = 3! = 6.
Total no. of arr. = 2 x 6 = 12.
Circle permutation
Q. In how many ways can 5 friends sit around a table if Bill and Ben do not sit next to each other?
Order important - circle permutation, with restriction.
Restricted items - Bill and Ben, unrestricted items - other 3 friends.
Restriction - Bill and Ben not next to each other,
Choose a location for Bill.
Arrange other four items as line permutation, with restriction.
Restriction Ben not on either end: F B F F
F F B F,
that is, 2 patterns.
Arr. of Ben (restricted item) = 1! = 1.
Arr. of 3 friends (unrestricted items) = 3! = 6.
Total no. of arr. = 2 x 1 x 6 = 12.
Combinations
Q. How many committees consisting of 3 teachers and 5 students can be made up if there are 6 teachers and 10 students to choose from?
Order unimportant - combination, with restriction.
Given: 6 teachers and 10 students.
Choose: 3 teachers and 5 students.
Comb. of teachers
Comb. of students
Total no. of committees = 20 x 252 = 5040.
Probability questions
Once students have mastered the different categories, solving probability questions becomes quite simple. Using the result: n(E) is an arrangement or
combination with restriction and n(S) is the same arrangement or combination without restriction. For example, refer to earlier examples to solve this question:
Q. 4 boys and 3 girls are arranged in a line. What is the probability that boys and girls alternate.
n(E) = 576 [See Q2, line perm., with restriction.]
n(S) = 5040 [See Q, line perm., without restriction.]
in item 1 on p.94.
Exceptions
There are some more challenging questions that do not fall into any of the categories given in Figure 1. In fact the gaps in Figure 1 suggest the types of questions that are not common to textbooks. For example, arranging some items of a word with repeated letters.
Q1.How many arrangements of three letters from the word BANANA are there?
3 possibilities - 3 unique letters, 3 letters with 2 identical, 3 letters all identical.
3 unique letters: 1 possibility - B, N, A:
Arr. = 3 ! = 6.
3 letters with 2 identical: 4 possibilities -
2 As with B or N.
2 Ns with B or A.
Arr.
Total = 4 x 3 = 12.
3 letters, all identical: 1 possibility - 3 As.
Arr.
Total no. of arr. = 6 + 12 + 1 = 19.
These questions are generally solved by considering individual cases. Others require calculation of a combination followed by a permutation. Try this one (from Fitzpatrick, p.194, Q23): How many words (arrangements of letters) containing 3 consonants and 2 vowels can be formed from the letters of the word promise? [Solution: 1440.]
Conclusion
With the exception of the charts, I have used this approach with two 3 Unit classes and it worked very well (much better than my first attempts). I first introduce them to the concepts of permutations and combinations (using a story about a hairdresser whose salon only offers perms and combs). We then look at the different categories, followed by an exercise where students provide the categories to which each question belongs. We next do problems in the same order as given in this paper. (Students are also shown the alternative methods of solution that exist for some types of questions.) This is followed by more theoretical aspects of the topic, which then leads into the binomial theorem. Strategies that help students with permutations and combinations include:
• using colours to distinguish the various sections of a solution;
• using students to model permutations and combinations;
• getting students to visualize the situation being described in the question;
• having printed lists of all possible arrangements of a word like SUNDAY to show that the large numbers are true;
• buying students a triple ice cream while they calculate how many hundreds of centuries it would take to produce all possible arrangements of 16 flavours if a new arrangement of tubs is set up each day.
The process of categorizing permutation and combination problems provides a methodical approach that helps both understanding and memory. Giving each category a label also helps some students. Overall, students definitely benefit from being able to approach questions systematically rather than haphazardly.