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Worksheet 4.11 Counting
Section 1 Introduction
When looking at situations involving counting it is often not practical to count things individ- ually. Instead techniques have been developed to help us count efficiently and accurately. In particular this chapter looks at permutations and combinations.
Firstly however we must look at the Fundamental Principle of Counting (sometimes referred to as the multiplication rule) which states:
If there are m ways of doing one thing and n ways of doing another, then there are m × n ways of doing the first thing followed by the second.
This rule can best be understood by looking at an example.
Example 1 : There are 3 t-shirts and 2 pairs of jeans in the cupboard. How many possible outfits are there? For each t-shirt there are 2 possibilities of jeans. All together we have three lots of these two possibilities, or, 3 × 2 = 6.
Example 2 : An ice-cream shop offers 3 types of cones and 5 different flavours of ice-cream. How many possible ice-cream cone combinations are there? For each of the 3 cones there are 5 possible toppings so altogether there are 3 × 5 = 15 possible ice-cream cone combinations.
Exercises:
- The local pizzeria offers a choice of 2 pizzas - supreme or vegetarian, 3 sides - chips, salad or coleslaw, and 4 drinks - juice, coke, ginger beer or water. For dinner I decide to have 1 pizza, 1 side, and 1 drink. How many possible meals do I have to choose from?
- How many different car number plates can be made if each is to display 3 letters followed by 3 numbers?
- Your friend wants to perform a magic trick and asks you to draw 2 cars from a standard deck of 52. The first card you draw must be placed face down and the second placed face up on the table. How many ways are there of drawing the 2 cards?
Section 2 Counting Techniques
For the many circumstances where we need to count the number of outcomes there are two different counting situations - permutations and combinations. A permutation is an arrange- ment where the order of selection matters. A combination is an arrangement where the order of selection doesn’t matter.
Example 1 : The number of ways of arranging 5 people in a line. There are 5 choices for the first spot, 4 choices for the second and so on, so we have 5× 4 × 3 × 2 ×1 = 5!. Therefore the number of ways of arranging n things into n places is n!.
Example 2 : To arrange 3 people out of 5 in a line there are 5 × 4 × 3 = 5!2! ways. These are all permutations, the number of ways of choosing k things out of n where the order matters is =
n! (n − k)!
= nPk.
How about choosing 3 out of 5 people to put on a committee? This time, it doesn’t matter whether you are chosen first, second or third, you are still on the committee. There are 5× 4 × 3 ways of choosing 3 people in order. However since order doesn’t matter we have over counted, and therefore we need to divide by the number of ways of arranging the three people that are chosen, 3!. So the number of ways of choosing this committee is the number of ways of choosing with order, divided by 3 × 2 × 1.
5 × 4 × 3 3 × 2 × 1
This is a combination, the number of ways of choosing k things from n where order does NOT matter
=
n! k!(n − k)!
= nCk.
Definition 1 : A permutation is the number of ways of choosing k things out of a possible n, where the order that they are chosen matters and is notated nPk.
nPk = n(n − 1)(n − 2)... (n − k + 1) = n! (n − k)!
Exercises:
- Calculate 10 P 4 and 10 C 4.
- Calculate 7 P 5 and 7 C 2.
- In an 18 team league, how many ways can the 8 teams for the finals be decided? In how many ways can the first 4 positions be decided?
- A child wants to draw a picture using only three different colours from a set containing twelve different colour pencils. In how many ways can the colours be chosen?
- In the front of a building there are three doors each to be painted a different colour from twelve different available colours. How many colour arrangements for the doors are there?
Section 3 Permutations
Example 1 :
(a) How many four digit numbers can be formed using only the digits 1, 2 , 3 , 4 , 5 , 6? (b) How many four digit numbers from (a) have no repeated digits? (c) How many four digit numbers from (b) are greater than 5000?
Answers:
(a) There are 6 possible digits for each of the four places in the number, so there are 6 × 6 × 6 × 6 = 6^4 = 1296 of these numbers. (b) There are 6 digits for the first place, and then only 5 digits for the second and so on. So there are 6 × 5 × 4 × 3 = 360 such numbers. (c) There are 2 choices for the first digit (a 5 or a 6), then 5 choices, 4 choices and 3 choices respectively for the remaining digits. So there are 2 × 5 × 4 × 3 = 120 such numbers.
Example 2 : Three adults and five children are seated randomly in a row.
(a) In how many ways can this be done? (b) In how many ways can this be done if the three adults are seated together? (c) In how many ways can this be done if the three adults are seated together and the five children are also seated together.
Answers:
(a) There are 8! ways of arranging 8 people in a row. (b) There are 3! ways of arranging the adults. We now need to arrange 6 objects (1 group of adults and 5 individual children) in a row. Therefore the answer is 3! × 6! ways. (c) There are 3! ways of arranging the adults, 5! ways of arranging the children, and 2! ways of arranging the 2 groups. So the answer is 3! × 5! × 2!.
Permutations with Repeated Objects
Example 3 : How many arrangements of the letters of the word IRRIGATION are there? Answer: There are 10 letters. If the letters were all different there would be 10! arrangements. However there are three I’s and two R’s and so we need to divide by 3! × 2!. Therefore the answer is
Example 4 : In how many ways can we rearrange the letters in “MATHS IS FUN”
(a) with no restrictions? (b) if the first and last letter must be vowels?
Answers:
(a) There are 10 letters to rearrange. Two of them are S’s and so that we do not over count we need to divide by the number of ways of arranging the S’, so there are
ways to arrange these letters.
(b) In the second instance, we first select the vowels and there are 3 × 2 ways of the selecting the first and last as vowels. Then we have 8 letters left with 2 S’s repeating so there are
ways to do this. Hence there are 3 × 2 ×
ways of arranging these letters so that the first and last are vowels.
Exercises:
- How many 4 digit numbers can be formed from the digits 1,2,3,4 if
(a) repetitions are allowed?
Exercises:
- In how many ways can you choose 2 chocolates from a bag containing 6 different choco- lates?
- Twelve dots are spaced equally around a circle. How many different triangles can be formed by joining dots?
- Tickets are numbered from 1 to 25. 6 tickets are chosen. In how many ways can this be done if the selection contains (a) all odd numbers? (b) 3 odd numbers and 3 even numbers? (c) the numbers 1 and 2?
Exercises for Worksheet 4.
- Decide whether or not order of selection is important and then calculate the following.
(a) How many different sets of three colours can be selected from the colours red, orange, yellow, green, blue, and violet? (b) In how many ways can a team of five basketball players be selected from 8 girls? (c) A race has 8 runners. In how many ways can the first three places be decided? (d) A secretary has nine letters and only five stamps. How many ways can he select the letters for posting?
- How many different possible “full house” (one pair, one three of a kind) hands are there in 5 card poker?
- We choose 12 cards from the usual deck of 52 playing cards.
(a) How many different ways can this be done? (b) How many ways can it be done if they must all come from the same suit? (c) How many ways can it be done if we need exactly 3 kings and 3 queens? (d) How many ways can it be done if all cards must have different face values?
- There are 10 boys and 11 girls at a school.
(a) How many different ways can the boys each choose a girl to take to the formal? (b) One of the girls doesn’t want to go to the formal, how many ways are there to make this choice now?
- A town of 30 people is to choose a committee of 3 to represent them, how many different ways can this be done? How many ways can it be done if one person is to be the chairperson, one the treasurer and one the secretary?
- In a certain electorate there are 6 candidates: labor, liberal, greens, and three indepen- dents. Their names are to be placed in random order on the ballot paper. In how many ways can this be done if
(a) the labor candidate comes first? (b) the liberal candidate comes first? (c) the three independent candidates are together?
- From a class of 30 students, five students are to be selected to complete a survey. In how many ways can the choice be made?
- In order to be photographed 10 people stand in two rows of 5, one in front of the other. In how many ways can this be done if
(a) there are no restrictions? (b) A and B are to be in the front row? (c) A and B are in different rows?
- In how many ways can you choose 10 people out of 30 to sit on a bench in order, if two particular people must be selected and seated together?
- You are to pick three singers for the Superbowl, one to sing before the game, one to sing at the end of the game, and one to sing at halftime. If there are 12 applicants, how many different possibilities are there if each singer can only sing once?
- Answers for Worksheet 4.
- Section
- Section
- Section
- (a) 256 (b)
- (a) 5040 (b) 240 (c) 1440 (d)
- (a) 360 (b) 20160 (c)
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