## 1. Introduction

The other day, I wanted to travel from Bangalore to Allahabad by train. There is no direct train from Bangalore to Allahabad, but there are trains from Bangalore to Itarsi and from Itarsi to Allahabad. From the railway timetable I found that there are two trains from Bangalore to Itarsi and three trains from Itarsi to Allahabad. Now, in how many ways can I travel from Bangalore to Allahabad?

There are counting problems which come under the branch of Mathematics called combinatorics.

In another situation suppose you are painting your house. If a particular shade or colour is not available, you may be able to create it by mixing different colours and shades. While creating new colours this way, the order of mixing is not important. It is the combination or choice of colours that determine the new colours; but not the order of mixing.

In this lesson we shall consider simple counting methods and use them in solving such simple counting problems.

### A. Useful Relations

(i). $0! = 1! = 1$

(ii). ${^nC_0} = \dfrac{n!}{0! \times n!} = \dfrac{1}{0!} = 1$

(iii). $n! = n \times (n-1)!$

(iv). $^nP_r = \dfrac{n!}{(n-r)!}$

(v). $(n+1). {^nP_r} = {^{n+1}P_{r+1}}$

(vi). ${^nC_r} = \dfrac{^nP_r}{r!}$

(vii). ${^nC_r} = {^nC_{n-r}}$

(viii). ${^{n-1}C_{r-1}} + {^{n-1}C_r} = {^nC_r}$

## 2. Fundamental Principle of Counting

Let us now solve the problem mentioned in the introduction. We will write $t_1, t_2$ to denote trains from Bangalore to Itarsi and $T_1, T_2, T_3$, for the trains from Itarsi to Allahabad. Suppose I take $t_1$ to travel from Bangalore to Itarsi. Then from Itarsi I can take$T_1$ or $T_2$ or $T_3$. So the possibilities are $t_1T_1, t_2T_2$ and $t_3T_3$ where $t_1T_1$ denotes travel from Bangalore to Itarsi by $t_1$ and travel from Itarsi to Allahabad by $T_1$. Similarly, if I take $t_2$ to travel from Bangalore to Itarsi, then the possibilities are $t_2T_1, t_2T_2$ and $t_2T_3$. Thus, in all there are $6(2 \times 3)$ possible ways of travelling from Bangalore to Allahabad.

Here we had a small number of trains and thus could list all possibilities. Had there been 10 trains from Bangalore to Itarsi and 15 trains from Itarsi to Allahabad, the task would have been very tedious. Here the Fundamental Principle of Counting or simply the Counting Principle comes in use:

If any event can occur in m ways and after it happens in any one of these ways, a second event can occur in n ways, then both the events together can occur in m \times n ways.

### Example 2.1

How many multiples of 5 are there from 10 to 95?

#### Solution:

As you know, multiples of 5 are integers having 0 or 5 in the digit to the extreme right (i.e. the unit's place).

The first digit from the right can be chosen in 2 ways.

The second digit can be any one of $1,2,3,4,5,6,7,8,9$

i.e. There are 9 choices for the second digit.

Thus, there are $2 \times 9 =$ 18 multiples of 5 from 10 to 95.

### Example 2.2

In a city, the bus route numbers consist of a natural number less than 100, followed by one of the letters A,B,C,D,E and F. How many different bus routes are possible?

#### Solution:

The number can be any one of the natural numbers from 1 to 99.

There are 99 choices for the number.

The letter can be chosen in 6 ways.

Number of possible bus routes are $99 \times 6 =$ 594

### Example 2.3

There are 3 questions in a question paper. If the questions have 4,3 and 2 solutions respectively, find the total number of solutions.

#### Solution:

Here question 1 has 4 solutions, question 2 has 3 solutions and question 3 has 2 solutions.

=> By the multiplication (counting) rule,

Total number of solutions$= 4 \times 3 \times 2 =$ 24

### Example 2.4

Consider the word ROTOR. Whichever way you read it, from left to right or from right to left, you get the same word. Such a word is known as palindrome. Find the maximum possible number of 5-letter palindromes.

#### Solution:

The first letter from the right can be chosen in 26 ways because there are 26 alphabets.

Having chosen this, the second letter can be chosen in 26 ways.

=> The first two letters can be chosen in $26 \times 26 = 676$ ways

Having chosen the first two letters, the third letter can be chosen in 26 ways.

=> All the three letters can be chosen in $676 \times 26 = 17576$ ways.

It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.

Note: In Example 2.4 we found the maximum possible number of five letter palindromes. There cannot be more than 17576. But this does not mean that there are 17576 palindromes. Because some of the choices like CCCCC may not be meaningful words in the English language.

### Example 2.5

How many 3-digit numbers can be formed with the digits 1,4,7,8 and 9 if the digits are not repeated?

#### Solution:

Three digit numbers will have unit's, ten's and hundred's place.

Out of 5 given digits any one can take the unit's place.

This can be done in 5 ways ------- (i)

After filling the unit's place, any of the four remaining digits can take the ten's place.

This can be done in 4 ways ------- (ii)

After filling in ten's place, hundred's place can be filled from any of the three remaining digits.

This can be done in 3 ways ------- (iii)

=> By counting principle, the number of 3 digit numbers $= 5 \times 4 \times 3 =$ 60

Let us now state the General Counting Principle

If there are n events and if the first event can occur in $m_1$ ways, the second event can occur in $m_2$ ways after the first event has occurred, the third event can occur in $m_3$ ways after the second event has occurred, and so on, then all the n events can occur in $m_1 \times m_2 \times ... \times m_{n-1} \times m_n$ ways.

### Example 2.6

Suppose you can travel from a place A to a place B by 3 buses, from place B to place C by 4 buses, from place C to place D by 2 buses and from place D to place E by 3 buses. In how many ways can you travel from A to E?

#### Solution:

The bus from A to B can be selected in 3 ways.

The bus from B to C can be selected in 4 ways.

The bus from C to D can be selected in 2 ways.

The bus from D to E can be selected in 3 ways.

So, by the General Counting Principle, one can travel from A to E in $3 \times 4 \times 2 \times 3 =$ 72 ways

## 3. Permutations

Suppose you want to arrange your books on a shelf. If you have only one book, there is only one way of arranging it. Suppose you have two books, one of History and one of Geography.

You can arrange the Geography and History books in two ways. Geography book first and the History book next, GH or History book first and Geography book next; HG. In other words, there are two arrangements of the two books.

Now, suppose you want to add a Mathematics book also to the shelf. After arranging History and Geography books in one of the two ways, say GH, you can put Mathematics book in one of the following ways: MGH, GMH or GHM. Similarly, corresponding to HG, you have three other ways of arranging the books. So, by the Counting Principle, you can arrange Mathematics, Geography and History books in $3 \times 2 = 6$ ways.

By permutation we mean an arrangement of objects in a particular order. In the above example, we were discussing the number of permutations of one book or two books.

In general, if you want to find the number of permutations of n objects $n \geq 1$, how can you do it? Let us see if we can find an answer to this.

Similar to what we saw in the case of books, there is one permutation of 1 object, $2 \times 1$ permutations of two objects and $3 \times 2 \times 1$ permutations of 3 objects. It may be that, there are $n \times (n-1) \times (n-2) \times ... \times 2 \times 1$ permutations of n objects. In fact, it is so, as you will see when we prove the following result.

### Theorem 3.1

The total number of permutations of n objects is $n (n - 1) ....2.1$

#### Proof:

We have to find the number of possible arrangements of n different objects.

The first place in an arrangement can be filled in n different ways. Once it has been done, the second place can be filled by any of the remaining $(n-1)$ objects and so this can be done in $(n-1)$ ways. Similarly, once the first two places have been filled, the third can be filled in $(n-2)$ ways and so on. The last place in the arrangement can be filled only in one way, because in this case we are left with only one object.

Using the counting principle, the total number of arrangements of n different objects is

$n(n-1)(n-2) ... 2.1$ ----------(3.1)

The product $n (n - 1) ... 2.1$ occurs so often in Mathematics that it deserves a name and notation. It is usually denoted by n! (read as n factorial).

$$n! = n (n - 1) ... 3.2.1$$

### Example 3.1

Suppose you want to arrange your English, Hindi, Mathematics, History, Geography and Science books on a shelf. In how many ways can you do it?

#### Solution:

We have to arrange 6 books.

The number of permutations of n objects is$n! = n. (n - 1) . (n - 2) ... 2.1$

Here $n = 6$ and therefore, number of permutations is $6.5.4.3.2.1 =$ 720

## 4. Permutation of r Objects Out of n Objects

Suppose you have five story books and you want to distribute one each to Asha, Akhtar and Jasvinder. In how many ways can you do it? You can give any one of the five books to Asha and after that you can give any one of the remaining four books to Akhtar. After that, you can give one of the remaining three books to Jasvinder. So, by the Counting Principle, you can distribute the books in $5 \times 4 \times 3$ i.e. 60 ways.

More generally, suppose you have to arranger objects out of n objects. In how many ways can you do it? Let us view this in the following way. Suppose you haven objects and you have to arrange r of these in r boxes, one object in each box.

| n ways|                    | n-1 ways |             .........               | n-r+1 ways |
____________________ r boxes________________________

Suppose there is one box. $r = 1$. You can put any of the n objects in it and this can be done in n ways. Suppose there are two boxes. $r = 2$. You can put any of the objects in the first box and after that the second box can be filled with any of the remaining $n - 1$ objects. So, by the counting principle, the two boxes can be filled in $n (n - 1)$ ways.

Similarly, 3 boxes can be filled in $n (n - 1) (n - 2)$ ways.

In general, we have the following theorem.

### Theorem 4.1

The number of permutations of $r$ objects out of $n$ objects is

$$n (n-1) .. (n - r + 1)$$

The number of permutations of $r$ objects out of $n$ objects is usually denoted by ${^nP_r}$.

Thus,${^nP_r} = n (n-1) (n-2) ... (n-r+1)$

#### Proof:

Suppose we have to arranger objects out of n different objects. In fact it is equivalent to filling r places, each with one of the objects out of the given n objects.

The first place can be filled in n different ways. Once this has been done, the second place can be filled by any one of the remaining $(n-1)$ objects, in $(n-1)$ ways.

Similarly, the third place can be filled in $(n - 2)$ ways and so on. The last place, the $r^{th}$ place can be filled in $[n-(r-1)]$ i.e. $(n-r+1)$ different ways. You may easily see, as to why this is so.

Using the Counting Principle, we get the required number of arrangements of $r$ out of $n$ objects

is $n (n-1) (n-2) ... (n-r+1)$

### Example 4.1

If you have 6 New Year greeting cards and you want to send them to 4 of your friends, in how many ways can this be done?

#### Solution:

We have to find number of permutations of 4 objects out of 6 objects.

This number is $^6P_4 = 6(6-1)(6-2)(6-3) = 6 \times 5 \times 4 \times 3 = 360$

Therefore, cards can be sent in 360 ways.

So, using the factorial notation, this formula can be written as follows:

${^nP_r} = \dfrac{n!}{(n-r)!}$

## 5. Permutations under Some Conditions

• Number of permutations of n different things, taken r at a time, when a particular thing is to be always included in each arrangement is: $r{^{n-1}P_{r-1}}$
•  Number of permutations of n different things, taken r at a time, when a particular thing is never taken in each arrangement is: ${^{n-1}P_r}$
•  Number of permutations of n different things, taken all at a time, when m specified things always come together is: $m! \times (n-m+1)!$
• Number of permutations of n different things, taken all at a time, when m specified never come together is: $n! - [m! \times (n-m+1)!]$
• The number of permutations of n dissimilar things taken r at a time when k(< r) particular things always occur is: $[{^{n-k}P_{r-k}}] \times [{^rP_k}]$
• The number of permutations of n dissimilar things taken r at a time when k particular things never occur is: $^{n-k}P_r$
• The number of permutations of n dissimilar things taken r at a time when repetition of things is allowed any number of times is: $n^r$
• The number of permutations of n different things, taken not more than r at a time, when each thing may occur any number of times is: $n + n^2 + n^3 + .... + n^r = \dfrac{n(n^r - 1)}{n-1}$
• The number of permutations of n different things taken not more than r at a time:
${^nP_1} + {^nP_2} + {^nP_3} + ... + {^nP_r}$

We will now see examples involving permutations with some extra conditions.

### Example 5.1

Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?

#### Solution:

Let the beds be numbered 1 to 7.

#### Case 1:

Suppose Anju is allotted bed number 1.

Then, Parvin cannot be allotted bed number 2.

So Parvin can be allotted a bed in 5 ways.

After allotting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.

So, in this case the beds can be allotted in $5 \times 5! = 600$ ways.

#### Case 2:

Anju is allotted bed number 7.

Then, Parvin cannot be allotted bed number 6

As in Case 1, the beds can be allotted in 600 ways.

#### Case 3:

Anju is allotted one of the beds numbered $2,3,4,5$ or $6$

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju's bed.

For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.

Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.

Therefore, in each of these cases, the beds can be allotted $4 \times 5! = 480$ ways.

=> The beds can be allotted in:

$2 \times 600 + 5 \times 480 = 1200 + 2400 =$ 3600 ways

### Example 5.2

In how many ways can an animal trainer arrange 5 lions and 4 tigers in a row so that no two lions are together?

#### Solution:

They have to be arranged in the following way:

| L | T | L | T | L | T | L | T | L |

The 5 lions should be arranged in the 5 places marked 'L'.

This can be done in 5! ways.

The 4 tigers should be in the 4 places marked 'T'.

This can be done in 4! ways.

Therefore, the lions and the tigers can be arranged in $5! \times 4! =$ 2880 ways

### Example 5.3

There are 4 books on fairy tales, 5 novels and 3 plays. In how many ways can you arrange these so that books on fairy tales are together, novels are together and plays are together and in the order, books on fairy tales, novels and plays.

#### Solution:

There are 4 books on fairy tales and they have to be put together.

They can be arranged in 4! ways.

Similarly, there are 5 novels.

They can be arranged in 5! ways.

And there are 3 plays.

They can be arranged in 3! ways.

So, by the counting principle all of them together can be arranged in $4! \times 5! \times 3! =$ 17280 ways

### Example 5.4

Suppose there are 4 books on fairy tales, 5 novels and 3 plays as in Example 5.3. They have to be arranged so that the books on fairy tales are together, novels are together and plays are together, but we no longer require that they should be in a specific order. In how many ways can this be done?

#### Solution:

First, we consider the books on fairy tales, novels and plays as single objects.

These three objects can be arranged in $3! = 6$ ways.

Let us fix one of these 6 arrangements.

This may give us a specific order, say, novels -> fairy tales -> plays.

Given this order, the books on the same subject can be arranged as follows.

The 4 books on fairy tales can be arranged among themselves in $4! = 24$ ways.

The 5 novels can be arranged in $5! = 120$ ways.

The 3 plays can be arranged in $3! = 6$ ways.

For a given order, the books can be arranged in $24 \times 120 \times 6 = 17280$ ways.

Therefore, for all the 6 possible orders the books can be arranged in $6 \times 17280 =$ 103680 ways.

### Example 5.5

In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together?

#### Solution:

Let 4 girls be one unit and now there are 6 units in all.

They can be arranged in 6! ways.

In each of these arrangements 4 girls can be arranged in 4! ways.

=> Total number of arrangements in which girls are always together

$= 6! \times 4! = 720 \times 24 =$ 17280

### Example 5.6

How many arrangements of the letters of the word ‘BENGALI’ can be made

(i) If the vowels are never together.

(ii) If the vowels are to occupy only odd places.

#### Solution:

There are 7 letters in the word ‘Bengali; of these 3 are vowels and 4 consonants.

(i) Considering vowels a, e, i as one letter, we can arrange 4+1 letters in 5! ways in each of which vowels are together. These 3 vowels can be arranged among themselves in 3! ways.

=> Total number of words $= 5! \times 3!$

$= 120 \times 6 = 720$

So there are total of 720 ways in which vowels are ALWAYS TOGEGHER.

Now,

Since there are no repeated letters, the total number of ways in which the letters of the word ‘BENGALI’ cab be arranged:

$=7! = 5040$

So,

$\text{Total no. of arrangements in which vowels are never together:}$

$= \text{ALL the arrangements possible - arrangements in which vowels are ALWAYS TOGETHER}$

$=5040 - 720= 4320$

(ii) There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in $^4P_3$ ways and 4 constants can be arranged in $^4P_4$ ways.

=> Number of words $= ^4P_3 \times ^4P_4 =$ 576

## 6. Combinations

Let us consider the example of shirts and trousers as stated in the introduction. There you have 4 sets of shirts and trousers and you want to take 2 sets with you while going on a trip. In how many ways can you do it?

Let us denote the sets by $S_1, S_2, S_3, S_4$. Then you can choose two pairs in the following ways:

1. {$S_1, S_2$} 2. {$S_1, S_3$} 3. {$S_1, S_4$}

4. {$S_2, S_3$} 5. {$S_2, S_4$} 6. {$S_3, S_4$}

[Observe that {$S_1, S_2$} is the same as {$S_2, S_1$}. So, there are 6 ways of choosing the two sets that you want to take with you. Of course, if you had 10 pairs and you wanted to take 7 pairs, it will be much more difficult to work out the number of pairs in this way.

Now as you may want to know the number of ways of wearing 2 out of 4 sets for two days, say Monday and Tuesday, and the order of wearing is also important to you. We know that it can be done in $^4P_4 = 12$ ways. But note that each choice of 2 sets gives us two ways of wearing 2 sets out of 4 sets as shown below:

1. {$S_1, S_2$} -> $S_1$ on Monday and $S_2$ on Tuesday or $S_2$ on Monday and $S_1$ on Tuesday

2. {$S_1, S_3$} -> $S_1$ on Monday and $S_3$ on Tuesday or $S_3$ on Monday and $S_1$ on Tuesday

3. {$S_1, S_4$} -> $S_1$ on Monday and $S_4$ on Tuesday or $S_4$ on Monday and $S_1$ on Tuesday

4. {$S_2, S_3$} -> $S_2$ on Monday and $S_3$ on Tuesday or $S_3$ on Monday and $S_2$ on Tuesday

5. {$S_2, S_4$} -> $S_2$ on Monday and $S_4$ on Tuesday or $S_4$ on Monday and $S_2$ on Tuesday

6. {$S_3, S_4$} -> $S_3$ on Monday and $S_4$ on Tuesday or $S_4$ on Monday and $S_3$ on Tuesday

Thus, there are 12 ways of wearing 2 out of 4 pairs.

This argument holds good in general as we can see from the following theorem.

### Theorem 6.1

Let $n \geq 1$ be an integer and $r \leq n$. Let us denote the number of ways of choosing $r$ objects out of $n$ objects by ${^nC_r}$. Then,

$^nC_r = \dfrac{^nP_r}{r!}$

### Example 6.1

Find the number of subsets of the set ${\{1,2,3,4,5,6,7,8,9,10,11\}}$ having 4 elements.

#### Solution:

Here the order of choosing the elements doesn't matter and this is a problem in combinations.

We have to find the number of ways of choosing 4 elements of this set which has 11 elements.

This can be done in:

${^{11}C_4} = \dfrac{11 \times 10 \times 9 \times 8}{1 \times 2 \times 3 \times 4} =$ 330 ways

### Example 6.2

12 points lie on a circle. How many cyclic quadrilaterals can be drawn by using these points?

#### Solution:

For any set of 4 points we get a cyclic quadrilateral. Number of ways of choosing 4 points out of 12 points is ${^{12}C_4} = 495$

Therefore, we can draw 495 quadrilaterals.

### Example 6.3

In a box, there are 5 black pens, 3 white pens and 4 red pens. In how many ways can 2 black pens, 2 white pens and 2 red pens can be chosen?

#### Solution:

Number of ways of choosing 2 black pens from 5 black pens

$= {^5C_2} = \dfrac{^5P_2}{2!} = \dfrac{5 \times 4}{1 \times 2} = 10$

Number of ways of choosing 2 white pens from 3 white pens

$= {^3C_2} = \dfrac{^3P_2}{2!} = \dfrac{3 \times 2}{1 \times 2} =3$

Number of ways of choosing 2 red pens from 4 red pens

$= {^4C_2} = \dfrac{^4P_2}{2!} = 4 \times \dfrac{3}{1 \times 2} = 6$

=> By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in

$10 \times 3 \times 6 =$ 180ways.

### Example 6.4

A question paper consists of 10 questions divided into two parts A and B. Each part contains five questions. A candidate is required to attempt six questions in all of which at least 2 should be from part A and at least 2 from part B. In how many ways can the candidate select the questions if he can answer all questions equally well?

#### Solution:

The candidate has to select six questions in all of which at least two should be from Part A and two should be from Part B. He can select questions in any of the following ways:

Table below can be scrolled horizontally

 Part A Part B (i) 2 4 (ii) 3 3 (iii) 4 2

If the candidate follows choice (i), the number of ways in which he can do so is:

${^5C_2} \times {^5C_4} = 10 \times 5 = 50$

If the candidate follows choice (ii), the number of ways in which he can do so is:

${^5C_3} \times {^5C_3} = 10 \times 10 = 100$

Similarly, if the candidate follows choice (iii), then the number of ways in which he can do so is:

${^5C_4} \times {^5C_2} = 5 \times 10 = 50$

Therefore, the candidate can select the question in $50 + 100 + 50 =$ 200 ways.

### Example 6.5

A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when:

(i) At least 2 women are included?

(ii) At most 2 women are included?

#### Solution:

(i) When at least 2 women are included.

The committee may consist of

3 women, 2 men: It can be done in ${^4C_3} \times {^6C_2}$ ways

Or, 4 women, 1 man: It can be done in ${^4C_4} \times {^6C_1}$ ways

or, 2 women, 3 men: It can be done in ${^4C_2} \times {^6C_3}$ ways

=> Total number of ways of forming the committee:

$= {^4C_3} \times {^6C_2} + {^4C_4} \times {^6C_1} + {^4C_2} \times {^6C_3} =$ 186 ways

(ii) When at most 2 women are included

The committee may consist of

2 women, 3 men: It can be done in ${^4C_2} \times {^6C_3}$ ways

Or, 1 women, 4 men: It can be done in ${^4C_1} \times {^6C_4}$ ways

Or, % men: It can be done in ${^6C_5}$ ways

=> Total number of ways of forming the committee:

$= {^4C_2} \times {^6C_3} + {^4C_1} \times {^6C_4} + {^6C_5} =$ 186 ways

### Example 6.6

The Indian Cricket team consists of 16 players. It includes 2 wicket keepers and 5 bowlers. In how many ways can a cricket eleven be selected if we have to select 1 wicket keeper and at least 4 bowlers?

#### Solution:

We are to choose 11 players including 1 wicket keeper and 4 bowlers

or, 1 wicket keeper and 5 bowlers.

Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players

$={^2C_1} \times {^5C_4} \times {^9C_6} = 840$

Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players

$={^2C_1} \times {^5C_5} \times {^9C_5} = 252$

=> Total number of ways of selecting the team:

$= 840 + 252 =$ 1092

## 7. Problems Involving Both Permutations and Combinations

So far, we have studied problems that involve either permutation alone or combination alone. In this section, we will consider some examples that need both of these concepts.

### Example 7.1

There are 5 novels and 4 biographies. In how many ways can 4 novels and 2 biographies can be arranged on a shelf?

#### Solution:

4 novels can be selected out of 5 in ${^5C_4}$ ways.

2 biographies can be selected out of4 in ${^4C_2}$ ways.

Number of ways of arranging novels and biographies:

$= {^5C_4} \times {^4C_2} = 30$

After selecting any 6 books (4 novels and 2 biographies) in one of the 30 ways, they can be arranged on the shelf in $6! = 720$ ways.

By the Counting Principle, the total number of arrangements $= 30 \times 720 =$ 21600

### Example 7.2

From 5 consonants and 4 vowels, how many words can be formed using 3 consonants and 2 vowels?

#### Solution:

From 5 consonants, 3 consonants can be selected in ${^5C_3}$ ways.

From 4 vowels, 2 vowels can be selected in ${^4C_2}$ ways.

Now with every selection, number of ways of arranging 5 letters is ${^5P_5}$

Total number of words $= {^5C_3} \times {^4C_2} \times {^5P_5} =$ 7200

## 8. Problem Solving Strategy

In permutation and combination problems, it is very important to recognize the type of problem. Many students mistakenly approach a combination problem as a permutation, and vice versa. The steps below will help you determine the problem type.

Solving a permutation or combination problem involves two steps:

1. Recognizing the problem type: permutation vs. combination.

2. Using formulas or models to count the possibilities.

We have three questions to ask ourselves in order to identify the problem type:

### 1. Is it a permutation or combination?

Check any two typical arrangements with the same combination. If the two arrangements are counted only once, it is a combination problem. Otherwise, it is a permutation.

#### Example:

If you are asked for a lock code, then 321 and 123 could be two possibilities, and the two numbers are formed from the same combination (Same number of 1’s, 2’s, and 3’s). So, lock codes must be permutations.

#### Example:

For another example, suppose you have 5 balls numbering 1 through 5. If you are asked to select 3 out of the 5 balls and you are only interested in the numbers on the balls, not the order in which they are taken, then you have a combination problem.

Problems that by definition connote ordering (though not directly stated) are permutations.

#### Example:

For example, 3 digits form a 3-digit number. Here, the 3-digit number connotes ordering.

#### Example:

For another example, if you are to answer 3 questions, you probably would not be asked to answer a particular question more than once. So, you would not allow repetition in the calculations. Though not often needed, such logical assumptions are allowed and sometimes expected.

If the problem itself defines slots for the arrangements, it is a permutation problem. Words like arrange" define slots for the arrangements.

Generally, "arrangements" refer to permutations, and "selections" refer to combinations. These words often flag the problem type.

Other words indicating permutations are "alteration," "shift," "transformation," and "transmutation," all of which connote ordering.

#### Example:

In how many ways can the letters of the word XYZ be transformed to form new words?

In how many ways can the letters of the word XYZ be altered to form new words?

Some words indicating combinations are "aggregation," "alliance," "association," "coalition," "composition," "confederation," "gang," "league," and "union," (all of which have nothing to do with arrangements but instead connote selections.)

#### Example:

In how many ways can a coalition of 2 countries be formed from 4 countries?

(Here, a coalition is the same whether you say country A and B are a coalition or country B and country A are a coalition.)

### 2. Are Repetitions Allowed?

Check whether, based on the problem description, the results of a permutation/combination can have repetitions.

#### Example:

If you are to list countries in a coalition, you can hardly list a country twice.

(Here, repetition automatically is not allowed unless specified otherwise.)

If you have 3 doors to a room, you could use the same door for both entering and exiting.

(Here, repetition is automatically allowed.)

### 3. Are There any Indistinguishable Objects in the Base Set?

Check the base set: the objects from which a permutation or a combination is drawn. If any indistinguishable objects (repetitions at base set level) are available, collect them. This is easy since it only requires finding identical objects in a base set, which is usually given.

#### Example:

If the original question is to find the words formed from the word GARGUNTUNG, then, in this step, you collect the information: G exists thrice; U exists twice, and so on.

Once the problem type is recognized, use the corresponding formula or model to solve it.

## 9. Let's Sum Up

Fundamental principle of counting states:

If there are $n$ events and if the first event can occur in $m_1$ ways, the second event can occur in $m_2$ ways after the first event has occurred, the third event can occur in $m_3$ ways after the second event has occurred and so on, then all the $n$ events can occur in:

$m_1 \times m_2 \times m_3 \times ... \times m_{n-1} \times m_n$ ways.

The number of permutations of n objects taken all at a time is $n!$

${^nP_r} = \dfrac{n!}{(n-r)!}$

${^nP_n} = n!$

The number of ways of selecting $r$ objects out of $n$ objects = ${^nC_r} = \dfrac{n!}{r! (n-r)!}$

${^nC_r} = {^nC_{n-r}}$

${^nC_r} + {^nC_{r-1}} = {^{n+1}C_r}$

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