Permutation, Combination, Probability, Slides of Probability and Statistics

Download Permutation, Combination, Probability and more Slides Probability and Statistics in PDF only on Docsity!

COMBINATIONS,

PERMUTATIONS AND

PROBABILITY

SECTION

Additional Mathematics^ Year 1

HANDLING DATA

Making Predictions with Data

INTRODUCTION

Permutations, Combinations and Probability are three fundamental concepts in mathematics. They have a wide range of practical applications in everyday life from organising events to predicting outcomes. These help us make informed decisions and solve complex problems. These topics, with roots tracing back to the 17th-century mathematicians Pierre de Fermat and Blaise Pascal, provide a powerful language for understanding and analysing chance and possibility in everyday life.

Have you ever wondered how many different ways you can arrange books on a shelf or make a selection from different fruits to eat?

Books on a shelve Banana Water melon Pawpaw

Figure 1: Items displayed to show different ways of arrangements or selection

What about selecting a team from a group of friends or determining the chances of winning a lottery? These are all scenarios where permutations, combinations and probability come into play. In this section, we will explore how to calculate these values and understand their significance through engaging examples. By the end, you will be able to apply these concepts to real-life problems, enhancing both your mathematical reasoning and decision-making skills.

At the end of this section, you will be able to:

- Use the fundamental counting principle to identify and determine the number of ways an event can occur - Understand the concept of permutation and combination and use it to solve related problems.

- Event (E) : A specific set of outcomes you are interested in. In a football penalty shootout (experiment), scoring a goal (outcome) is an event you might be interested in. - Outcome: The specific result of a single trial in an experiment. Each kick in the penalty shootout (experiment) has a specific outcome - goal (success) or miss (failure). - Sample Space (S) : The collection of all possible outcomes for an experiment. For tossing a coin, the sample space is {heads, tails}. At a market stall selling apples, oranges and pineapples (experiment), the sample space is {apple, orange, pineapple}. - Mutually exclusive events: Two events A and B are said to be mutually exclusive if event A and event B cannot happen simultaneously - Independent events: Event A is said to be independent of event B if the occurrence of A does not affect the probability of the B

FUNDAMENTAL COUNTING PRINCIPLE

(MULTIPLICATION RULE)

Figure 2: Athletes running a race

Look at the picture, how were the athletes arranged in the lanes fairly for the competition? This and many other related questions on how arrangement and selections are done will be the focus. The concept is applied in:

1. Menu and recipe planning 2. scheduling and time management 3. voting systems 4. password and security 5. genetics

Application of fundamental principles of counting

Ordered and unordered arrangements

How do you arrange your books on your shelf? How do you arrange your clothes in your wardrobe? The answer lies in counting rules.

Consider the Table 1 below, which represents the tossing of coin (H/T) and die (1, 2, 3, 4, 5, 6) once.

Example 1

How many possible outcomes are there if a coin (heads/tails) is flipped and a card, numbered from 1 to 10 is selected.

Solution

2 (coin) × 10 (numbered cards) = 20 combinations.

Example 2

Three SHS1 students, Naa, Afiba and Malik, are to be photographed. How many different ways are there to arrange them on 3 chairs?

Solution

Let Naa be represented by N, Afiba by A and Malik by M.

Table 3: Probability Experiment-Arranging chairs

Arrangements First Chair Second Chair Third Chair 1 N A M (^2) N M A (^3) A M N (^4) A N M (^5) M A N (^6) M N A

There are 6 possible arrangements.

Note, that in an arrangement order matters. If you were simply selecting them for the photo then there is only 1 way as order does not matter.

Example 3

You took three books Mathematics (M), English Language (E) and Home Economics (H) for prep. In how many ways can the subjects be:

a. arranged b. be selected

Solution a. for arrangement, it means order matters.

MEH

MHE EMH EHM HME HEM

The books can be arranged in 6 ways. b. for selection order is not important.

MEH

The books can be selected in 1 way.

In the above examples, where the order of events was important, we use permutations and where order was not important we use combinations. This gives us two forms of arrangements: ordered arrangement and unordered arrangements.

Ordered Arrangements

When the order of events matter, we use permutations. Imagine arranging four friends in a line for a photo. Each person has different positions they can occupy. The number of arrangements, considering order, is calculated using factorials ( n!

). For four ( 4 ) friends, there would be 4! = 24 unique orderings. We can also use

the permutation button on our calculators nPr.

Unordered Arrangements

If the order doesn’t matter, we use combinations. For example, choosing a fruit salad from a variety of fruits. Whether you pick an apple first or a banana doesn’t change the combination. Combinations are calculated using a formula involving factorials and selections and we can also use the combination button on our calculators, n C r.

Example 4

Four athletes, A, B, C, D are to be arranged in the four lanes (1, 2, 3, 4) for a competition.

Figure 4: A Running Track with 4 lanes

In how many ways can this be done?

Solution

If A goes first, he or she has 4 lanes available

B will have 3 lanes because A has occupied one already

C will have 2 lanes available

D will have 1 lane available

Multiplying all the lanes available to them, we obtain

4 × 3 × 2 × 1 = 4 != 24

Example 5

A girl has three different skirts (P, V, R) and 5 different tops (B, W, I, Y, L). If she can wear each skirt with each top, how many different outfits could she make?

Solution

One way is to use table to do the pairing

Table 4: Probability Experiment-Making different outfits from skirts and tops

B W I Y L P (^) P,B P,W P,I P,Y P,L V (^) V,B V,W V,I V,Y V,L R (^) R,B R,W R,I R,Y R,L

This shows us that there are 15 ways

Alternatively, we could multiply the number of skirts by the number of tops.

3 × 5 = 15

Example 6

In how many ways can you select 3 books from 7 different books.

Solution

Step 1 : Out of 7 books, the first book can be selected in 7 ways

Step 2 : The second book can be selected 6 ways for the remaining 6 books

Step 3 : The third book can be selected 5 ways for the remaining 5 books

Therefore, selecting 3 books out of the 7 books can be done in 7 × 6 × 5 = 210 different ways.

Example 7

How many numbers of 3 different digits can be formed with no repetition by choosing from the digits 1, 2, 3, 4 and 5?

Solution

Step 1 : choosing the hundred’s digit: we can do this in 5 ways

Step 2 : choosing the ten’s digit: we can do this in 4 ways since one digit is fixed as a hundred digit.

Permutations

The general rule for the number of permutations of n different items taking all at once is n! Before we solve more examples on finding permutation of objects, let us revise the concept of factorial, n!. We were introduced to the concept of factorial notation under binomial theorem where it was observed that n != n ( n − 1)( n − 2)( n − 3)…….

Don’t forget that 0 != 1

Example 9

Evaluate the following and verify your answers using your calculator

a. 3! b. 6! c. 5! d. 8! _ 6! e. 7! _ 9!

Solution a. 3! = 3(3 − 1)(3 − 2) = 3 × 2 × 1 = 6 b. 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 c. 5! = 5 × 4 × 3 × 2 × 1 = 120 d. __^86!! = _____________________8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 6 × 5 × 4 × 3 × 2 × 1 = 8 × 7 = 56

e. __^79!! = ________________________ 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 17 × 6 × 5 × 4 × 3 × 2 × 1 = __ 721

Example 10

Evaluate:

a. ______ ( n − 1) n^!! b. ______( ( nn^ + 1− 1))!! c. ______ ( n + 2) n^!! d. ______( ( nn^ − 1)− 2)^ !!

Solution

a. ______ ( n − 1 n^! )! = _____________ n (^ n ( n − 1)^ − 1)( n ( n − 2)^ − 2)…… = n

b. ______( ( nn^ + 1− 1))!! = __________________( n^ + 1) ( n − 1) n ( n^ − 1)( n − 2)( n^ − 2)… …= ( n + 1) n

c. ______ ( n + 2 n^! )! = _______________________ ( n + 2) n ( n ( n + 1)^ − 1) n (( nn^ − 2)− 1)…( n − 2)… = ___________ ( n + 2 _^1 ( n + 1)

d. ______( ( nn^ − 1− 2))!! = _________________( n^ − 1) ( n − 2)( n^ − 2)( n − 3)( n^ − 3)… …= n − 1

Now let us use this in permutations:

Example 11

1. Find the number of permutations or unique arrangements of the letters in the following words a. AYO b. OWARE c. CANOE

Solution a. AYO = 3! = 3 × 2 × 1 = 6 b. OWARE = 5! = 5 × 4 × 3 × 2 × 1 = 120 c. CANOES = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Example 12

Kweku has 4 different mathematic books on his shelf. Find the number of ways in which the 4 books can be arranged on the shelf.

Solution There will be 4 choices in the first slot There will be 3 choices in the first slot There will be 2 choices in the first slot There will be 1choices in the first slot ∴ There are 4 × 3 × 2 × 1= 4! = 24 permutations

We can use our calculators for finding permutations, with the nPr button. Use your calculator to evaluate 6 P 2. Hopefully you got 30. What about 8 P 3? The answer is

If you don’t have a calculator, we can compute this manually.

The number of ways in which r items can be selected out of a set n items in which order of arrangement matters is given by the formula:

nP r =^ P ( n ,^ r ) =^

______ n^! ( n − r )!

Example 15

Find the number of permutations (unique arrangements) of the letters in the following words

a. HOME b. ADOMI c. CASTLE d. STUDENT e. CHEMISTRY

Solution a. HOME =^4 P 4 = 4 != 4 × 3 × 2 × 1 = 24 ways b. ADOMI =^5 P 5 = 5 != 5 × 4 × 3 × 2 × 1 = 120 ways c. CASTLE =^6 P 6 = 6 != 6 × 5 × 4 × 3 × 2 × 1 = 720 ways d. STUDENT =^7 P 7 = 5 != 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 ways e. CHEMISTRY =^9 P 9 = 9 != 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362880 ways

Example 16

How many ways can 6 books be arranged out of 15 books?

Solution

(^15) P 6 =^

_______^15! (15 − 6)! =^

___^15! 9! = ______________________________________________15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3 603 600 ways

Permutations with identical/repeated objects

What happens if we want to find permutations where some object or letter is repeated? For example, in the word BOOKS, the O is repeated. In these cases we find the permutation and divide it by the product of the number of identical objects:

_______________^ n^! ( r 1 !. r 2 !. r 3 !…^ .. rk !)

Example 17

In how many ways can the letters of the word STATISTICS be arranged?

Solution

The word STATISTICS has 10 letters with 3 S s , 3 T s and 2 I s

Thus the numbers ways for arranging the letters in the word STATISTICS is

________^10! 3 !× 3 !× 2! =^ 50 400^ ways

Example 18

How many ways can you arrange the letters in the following words

a. COMMISSION b. SYLLABUS c. CURRICULUM d. BUSINESS

Solution a. COMMISSION = 10 letters with 2Os, 2Ms, 2Is, 2Ss ___________^10! 2! × 2! × 2! × 2! = 226 800^ ways b. SYLLABUS = 8 letters with 2Ss, 2Ls _____^8! 2! × 2! = 10 080^ ways c. CURRICULUM = 10 letters with 2Cs , 3Us, 2Rs ________^10! 2! × 3! × 2! = 151 200^ ways d. BUSINESS = 8 letters with 3Ss __^8! 3! = 6 720^ ways

Solution a. If the O’s must come together, then the word may be LK(OO) LK(OO) = 3 letters = 3! = 6 arrangements Number of arrangements of the Os (OO), 2 letters with 2 letters repeating _^2! 2! = 1 Multiplying the two results, 6 × 1 = 6 arrangements b. The O’s must be separated. Find the numbers of ways of arranging LOOK = 4 letters with 2 letters repeating _^4! 2! = 12 Subtract the result of when the Os came together from the arrangement of the word LOOK 12 − 6 = 6 arrangements

Example 21

In how many ways can the letters in the word BAILIFF be arranged if

i. The Fs must come together. ii. The Fs must be separated.

Solution i. If the F’s must come together, then the word may be BAIIL(FF) BAIIL(FF) = 6 letters with the Is repeating = __^62!! = 360 arrangements

Arranging (FF) which has 2 letters with F’s repeating: __^22!! = 1

Hence if the F’s must come together, then 360 × 1 = 360 arrangements. ii. The word BAILIFF has 7 letters with 2Is and 2Fs _____^7! 2 !× 2! = 1260^ arrangements. With the Fs separated = 1260 – 360 = 900 arrangements

Cyclic Permutations

Figure 5: Circular table with chairs

What happens if we want to arrange four people around a circular table? The circular table has no end, so one person will sit at a place (fixed) leaving 3 people for the arrangement on the three remaining seats. This can be written as 1 × (4 − 1)! = 1 × 3 = 3 ways.

What if we have 7 chairs around the circular table? 1 person will be fixed and the rest be arranged. i.e. 1 × (7 − 1)!= 1 × 6 != 720 ways.

So, in general, given n objects to arranged in a circle is given as

1 × ( n − 1)! = ( n − 1)!

Example 22

Six prefects attend a meeting in a conference room with a round table. How many ways can they arrange themselves for the meeting?

Solution

(6 − 1)! = 5! = 5 × 4 × 3 × 2 × 1 = 720 ways

Example 23

10 girls formed a circle to play ampe. How ways can they arrange themselves?

(10 − 1)! = 9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362 880 ways