H2 Mathematics Tutorial PROB1 – Permutations and Combinations

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H2 Mathematics Tutorial PROB1 – Permutations and Combinations

Questions for Extra Discussion (QED) Solutions QED will be discussed in tutorial lessons. Please attempt QED and complete it as much as

possible before coming for class so that you can contribute to the discussion.

1 Certain counting questions can be difficult to solve directly by elementary methods,

due to some conflicts which occur when trying to enumerate directly. In these types of

questions, we have to divide our problem into various mutually exclusive and

exhaustive cases such that each case is easy to enumerate.

(a) Find the number of ways to select five people from seven boys and three girls

such that at least four boys are selected.

(b) Find the number of even five digit numbers less than 50000 with non-repeating

digits.

(c) Find the number of ways to arrange seven boys and three girls in a row such that

there are at least three boys between any two girls.

(d) Find the number of ways to select five different numbers from 1 to 25 such that

their sum is odd.

(e) Find the number of different ways to choose four letters from the letters in

“MISSISSIPPI”.

Solution

(a) Number of ways

.

(Case 1: four boys, Case 2: five boys)

(b) Number of ways . (Case 1: first digit odd, Case 2: first digit even) (c) Number of ways . (BGBBBGBBBG, GBBBBGBBBG, GBBBGBBBBG, GBBBGBBBGB)

(d) Number of ways

.

(Case 1: 5 odd, Case 2: 3 odd, 2 even, Case 3: 1 odd, 4 even)

(e) Number of ways

.

(Case 1: 4 identical, Case 2: 3 identical 1 distinct, Case 3: 2 identical 2 identical, Case 4: 2 identical, 2 distinct, Case 5: 4 distinct)

2 If a certain counting problem appears too tedious to do directly, we should try to

approach the problem by taking complements.

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(a) Find the number of ways to select five people from seven boys and eight girls

such that at least one boy is selected. Try to do this question in two different

ways:

(i) By dividing into four cases of one, two, three and four selected boys;

(ii) By using complements: consider the total number of ways to select five

people from 15 people, and the special case where no boys are selected.

Which method do you prefer and why?

(b) (i) Find the number of ways to arrange the ten different objects in a row such

that two particular objects are not beside each other.

(ii) Find the number of ways to arrange the letters in “MISSISSIPPI” such that

the two “P”s are not beside each other.

(c) A focus group consists of ten people, three of whom are John, Mark and Peter.

How many ways are there to divide the ten people into three groups of 2, 3, and

5 people respectively such that John, Mark and Peter are not all in the same

group?

(d) From 1 Primary One boy, 2 Primary Two boys, 3 Primary Three boys, 4 Primary

Four boys and 5 Primary Five boys, how many ways are there to

(i) seat them in a row such that the Primary Three boys are all seated

together but the Primary Four boys are not all seated together?

(ii) select seven boys such that the boys come from at least three different

levels?

Solution

(a) (i) Number of ways

.

(ii) Number of ways

.

(b) (i) Number of ways .

(ii) Number of ways

.

(c) Number of ways

.

(Total minus all in group of 3, all in group of 5)

(d) (i) Number of ways . (No of ways such that P3 boys seated together, minus no of ways such that P3 boys seated together and P4 boys seated together)

(ii) Number of ways

.

(Total minus Two/Five, Three/Four, Three/Five and Four/Five)

3 When we face a counting problem where we require a category of objects to all be

separated from each other, the “slotting in” method described below becomes useful.

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(a) In a group of 7 boys and 4 girls, we want to arrange them in a row such that no

girl is next to another girl. We first arrange the 7 boys in a row. After that, the four

girls are assigned to (“slotted into”) 4 of the 8 possible spaces between the boys

(including the two ends). How many ways are there to perform this arrangement?

(b) In a group of 7 boys and 3 girls, calculate the number of ways to arrange them in

a row such that no girl is next to another girl.

(c) Find the number of ways to arrange the letters in “MISSISSIPPI” such that no

two “S”s are adjacent.

(d) In a group of 7 boys and 4 girls, calculate the number of ways to arrange them in

a circle such that no girl is next to another girl.

(e) In a group of 6 boys, 4 girls and 3 teachers, calculate the number of ways to

arrange them in a row such that there is at least a boy between any two girls,

and no teacher is next to another teacher.

Solution

(a) Number of ways .

(b) Number of ways .

(c) Number of ways

. (or

)

(d) Number of ways ( ) .

(e) Number of ways

.

4 This question introduces two different ways to deal with more complex permutations

involving repeated / identical objects, both of which have their pros and cons.

(a) To calculate the number of ways of arranging the letters in “MISSISSIPPI” in a

row such that the word “MISS” appears in the row, try the following two methods

(i) Try forming “MISS” as one object (only one way to do this), and consider

the arrangement of this object with the other letters. Be careful to divide by

the correct factorials when doing the arrangement!

(ii) Try treating all repeated letters as distinct (i.e. ),

count the number of ways to form the object “MISS” (no longer just one

way), consider the arrangement of this object with the other letters, and

finally dividing by the respective factorials to account for the overall

identical letters.

These two methods should yield the same result. Method (i) results in shorter

working, but may result in some confusion with regards to what factorials to

divide by. Method (ii) results in longer working and higher potential of careless

mistakes, but is generally more procedural in nature and more systematic to

execute.

(b) Given 2 identical red balls, 3 identical blue balls and 4 identical green balls, find

the number of ways to arrange the balls in a row such that

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(i) all the blue balls are together,

(ii) the first three balls in the row are of three different colours.

(iii) the first three balls in the row are of three different colours and the last

three balls are not of three different colours

(c) Calculate the number of ways of arranging the letters in “MISSISSIPPI” in a row

such that two of the words “ISS” appear in the row.

Solution

(a) Number of ways

. (or (

)

)

(b) (i) Number of ways

. (or

)

(ii) Number of ways

. (or

)

(iii) Number of ways

.

(or

)

(c) Number of ways

. (or (

)(

)

)

5 To deal with circular permutations involving numbered / distinct seats, we perform the

permutation by first considering that the seats are identical, then by multiplying with the

number of seats to ensure that all different perspectives are considered.

(a) Find the number of ways to arrange 7 boys and 3 girls around a circular table

with numbered seats. Is this the same as the number of ways to arrange them in

a row? Why?

(b) Find the number of ways to arrange 7 boys and 3 girls around a circular table

with numbered seats, such that the 3 girls are seated together. Is this the same

as the number of ways to arrange them in a row such that the 3 girls are seated

together? Why?

Solution

(a) Number of ways ( ) . There is a one-one correspondence of each numbered seat with a row position (b) Number of ways ( )

In a row: Number of ways In the circle, we allow the 3 girls to sit in seated numbered, for eg 9, 10, 1, as they will still be beside each other, but this will not be allowed in a row.

6 When dividing distinct objects into unlabeled groups of identical sizes, remember to

divide by the respective factorials to eliminate double counting situations.

(a) How many ways are there to arrange 12 people into

(i) Two groups of 6 people each,

(ii) One group of 2 people, and two groups of 5 people each,

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(iii) Three groups of 4 people each,

(iv) Two group of 3 people each, and three groups of 2 people each.

(b) Suppose among the 12 people involved, two of them are John and Peter. Find

the number of ways to arrange these 12 people into the following group

configurations, such that John and Peter are in separate groups:

(i) Two groups of 6 people each,

(ii) Four groups of 3 people each,

(iii) One group of 2 people, and two groups of 5 people each.

What fundamental difference do you observe, and why?

(c) How many ways are there to place 12 different objects into 4 identical bags, such

that each bag has at most 4 objects? Bags can be left empty.

Solution

(a) (i) Number of ways

.

(ii) Number of ways

.

(iii) Number of ways

.

(iv) Number of ways

.

(b) (i) Number of ways

.

(ii) Number of ways

.

(iii) Number of ways

. The groups with John and Peter within are considered as two distinct groups, even if they have the same number of people, since the members John and Peter identifies each group. This plays a role when considering the factorials we divide by.

(c) Number of ways

.

7 Usually for circular permutations, where two particular people are to be seated apart

by people, it is a relatively simple matter of forming a particular cluster with the two

people at the ends and others in between, then calculating the circular permutation

of that cluster with everybody else. However, over-counting may occur when the two

particular people are diagonally opposite each other.

(a) There are 10 people to be seated in a circle, two of whom are John and Peter.

How many ways are there to do so such that John and Peter are separated by

exactly two other people?

(b) Consider the same situation of ten people (including John and Peter) to be

seated in a circle, such that John and Peter are separated by exactly four other

people. Try to perform this counting by the following three methods:

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(i) By fixing Peter at one particular spot, thereby fixing John diagonally

opposite, and the eight remaining people to be arranged in the eight

spaces around Peter.

(ii) By using a similar idea in (a).

What did you notice that you need to do for (ii)? Why?

(c) There are now 12 people to be seated in a circle, including John, Peter and Mary.

How many ways are there to do so such that John, Peter and Mary are each

separated by exactly three other people? Try to do it using both perspectives!

Solution

(a) Number of ways ( ) .

(b) (i) Number of ways .

(ii) Number of ways ( )

.

We need to divide by for (ii). Suppose if the cluster is chosen as “J1234P” and wrap “5678” around, we would get the same combination if we chose the cluster “P5678J” and wrap “1234” around. Observing as such, there is a double counting situation involved.

(c) Number of ways .

8 There are many other situations like the above where over-counting could occur. We

must always ask ourselves if the technique we have used would lead us to over-

counting.

We would like to calculate the number of ways to choose 5 people out of 8 married

couples such that no married couple is chosen together. Refer to the following parts:

(i) A student considers that there are 16 ways to choose the first person, 14 ways to

choose the second, 12, 10 and 8 ways to choose the third, fourth and fifth

persons respectively, hence concluding that there are

ways to perform the selection. What’s wrong? How could we fix the method of

counting?

(ii) Can you think of other ways to perform this counting task?

Solution

(i) He forgot that he’s choosing 5 people without considering their order within the selection, so there is no so-called first, second, etc persons. He could fix this by dividing by to get rid of the order of selection, so that over-counting

doesn’t occur. Hence, number of words is

.

(ii) Choose 5 couples first, then choose 1 member of each couple. Number of

ways .

9 There are certain counting situations for which splitting into cases seem ideal.

However, if we try to do it we realize that the cases cannot be “cleanly” divided: there

will be some overlap between the various cases. We can see that the number of ways

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to solve the original problem is not simply the sum of the number of ways to count

each of the individual cases: over-counting would occur.

(a) (i) How many integers from 1 to 1000 (inclusive) are divisible by 3?

(ii) How many integers from 1 to 1000 (inclusive) are divisible by 5?

(iii) How many integers from 1 to 1000 (inclusive) are divisible by both 3 and 5,

i.e., divisible by 15?

(iv) Use (i), (ii) and (iii) to find the number of integers from 1 to 1000 (inclusive)

are divisible by either 3 or 5, or both.

(v) How many integers from 1 to 1000 (inclusive) are neither divisible by 3 nor

5?

Use the idea behind the above method to attempt the following problems:

(b) How many integers from 1 to 1000000 are perfect squares or perfect cubes (or

both)? (Note that )

(c) How many ways are there of arranging three Primary One students, three

Primary Two students, two Primary Three students and two Primary Four

students in a circle such that no three adjacent students are from the same level?

(d) (i) How many ways are there of arranging the letters in MISSISSIPPI such

that either the word “MISS” or the word “PISS” is present, or both?

(ii) How many ways are there of arranging the letters in MISSISSIPPI such

that either the word “MISS” or the word “SIPS” is present, or both? Note

that there is a key complication in this counting situation compared to (i)

that we need to take note of.

(e) [LOST (Learn Out of Syllabus Things)] How many ways are there of arranging

the letters in MISSISSIPPI such that none of the words “MISS”, “SIPS” or “SIS”

appear?

Solution

(a) (i) Number of integers

(ii) Number of integers

(iii) Number of integers

(iv) Number of integers (v) Number of integers (b) Number of integers . (c) Number of ways ( ) ( ( ) ( ) ( ) ) .

(d) (i) Number of ways

. (case with MISS, case with PISS,

minus case with both MISS and PISS)

(ii) Number of ways

(

) . (Taking into account both

MISS/SIPS and MISSIPS for the overlapping case)

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(e) Number of ways

(

((

) (

)

(

)) ( )) .

10 [LOST (Learn Out of Syllabus Things)] Many beautiful combinatorial identities can

be proven by simply solving counting problems in two different ways.

(a) By considering the number of ways to select any number of objects from

objects, show that

.

Think of various counting situations that can be used to prove the following

combinatorial identities:

(b)

(c)

,

(d)

, where

(e) ( ) (

) ( ) (

)

Solution

(a) LHS is obtained by considering the cases where 0 object is chosen, 1 object is

chosen, 2 objects are chosen, …, objects are chosen. RHS is obtained by determining, for each object, if it would be chosen.

(b) From objects, selecting objects, versus NOT selecting objects. (c) From objects, select objects. RHS is obtained by considering if a particular object is chosen, and the two cases that follows.

(d) From objects, select a shortlist of objects, then choose from these objects. RHS is obtained by first selecting the objects, then from the rest, selecting the other objects that made it into the shortlist. (e) From objects, choose . LHS is obtained by considering the different cases: from

the first objects, choose , and from the next objects, choose , with . Use identity in (b) as well.