Permutations and Combinations - Counting Techniques

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This module is very important in regards of basic counting skills. An aspirant should learn this skill for the further effective applications of the concepts of Permutation and Combination. In this module we are not dealing with any kind of ever remembering formulae or equations. It is purely based on your logical counting skills. So whenever you are approaching such a problem, never go for any kind of per-defined equation or formulae, just think about the logical trick behind each question.

In this article on counting techniques, you saw the different approaches of logical counting methods. Technically that part is called the "Basic Counting Principles".

The main objectives coming under this module are;
  • Basic Counting Principles
  • Fundamental rules of Addition and Multiplication.
  • Concept of arrangements (with repetition of objects allowable)
  • Conditional arrangements

Basic Counting Principles

This is the foundation concept of the logical counting, hence a plethora of questions that we can expect from this concept in various exams. Here we are not using any kind of pre-defined results for solving the problems under this requirement. You have to use your logical sense only for tackling these questions. Without any further explanations, we can move to the questions to check your logical counting skills.

Sample questions:

What is the total number of key depressions required to type the numbers from 1 to 400, without leaving any space in-between, on a computer monitor?

A. 80200

B. 400

C. 1092

D. 5050

Solution:

We can classify the numbers into three groups, such as, single digit numbers, two digits numbers and three digit numbers.

Range of numbersNumber of numbersRequired key-depressions for each numberTotal number of key-depressions required
1 to 9 9 1 9 * 1 = 9
10 to 99 90 2 90 * 2 = 180
100 to 400 301 3 301 * 3 = 903

Hence the total number of key-depressions = 9 + 180 + 903 = 1092

Ans: C

How many distinct sets of 5 consecutive natural numbers can be formed from the first 50 natural numbers, such that each of the sets should contain at least one multiple of 5 ?

A. 46

B. 50

C. 35

D. 40

Solution:

The required sets are mentioned below.

{1, 2, 3, 4, 5}

{2, 3, 4, 5, 6}

{3, 4, 5, 6, 7}
.
.

{46, 47, 48, 49 50}

From the above illustration, it is easy to identify that there are five different sets are possible by including '5'.

Similarly such five different sets are possible by including each of the further multiples of 5 up to 45, but only one set is possible by including 50.

Hence the total number of possible distinct sets = (5 * 9) + 1 = 46

Ans: A

Find the total number of all squares on a chess board?

A. 64

B. 127

C. 254

D. 204

Solution:

If we observe this chessboard, there are different types of squares lying on the chessboard.

1 * 1 squares, 2 * 2 squares,..., 8 * 8squares

Total number of 1 * 1 squares = 8 * 8 = 82

[There are eight 1 * 1 squares on each horizontal and vertical rows]

Similarly, total number of 2 * 2 squares = 7 * 7 = 72

Total number of 3 * 3 squares = 6 * 6 = 62
.
.

Total number of 8 * 8 squares = 1 * 1 = 12

Total number of squares on a chess board = 12 + 22 + 32 + 42 + 52 + 62 + 72 + 82

= (8 * 9 * 17)/6 (i.e. sum of first 8 natural squares)

= 204

Ans: D

Generalization of the concept:

Total number of squares in an 8 * 8 chess board = 12 + 22 + 32 + 42 + 52 + 62 + 72 + 82

Total number of squares in a 9 * 9 square board = 12 + 22 + 32 + 42 + 52 + 62 + 72 + 82 + 92

Total number of squares in an N * N square board = 12 + 22 + 32 + 42+ ..... + N2

Or in other words, total number of squares in a board which is formed by intersecting 'n' horizontal and 'n' vertical lines = 12 + 22 + 32 + 42+ ...... + (n - 1)2

Find the total number of squares in a 10 * 10 square board.

Solution:

Number of squares = 12 + 22 + 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102

= 385

Number of particular dimension squares in a square board

Total number of m * m squares in an n * n board, (where n > m) = (n - m + 1)2

Since there are n+1 horizontal and vertical lines in an n * n square board.

Examples:

Total number of 3 * 3 squares on an chess board = (8 - 3 + 1)2 = 62 = 36

Total number of 4 * 4 squares on an chess board = (8 - 4 + 1)2 = 52 = 25

Find the total number of 4 * 4 squares in a 20 * 20 square board.

  • 256
  • 289
  • 324
  • 361

Solution:

Number of 4 * 4 squares = (21 - 4)2 = 172 = 289

Ans: B

What is the total number of 10 * 10 squares in a 50 * 50 square board?

  • 1600
  • 1521
  • 1681
  • 1764

Solution:

Number of 10 * 10 squares = (51 - 10)2 = 412 = 1681

Ans: C

Total number of squares in an m * n board

Example:

Total number of squares in an 8 * 9 board.
In this case, the board is not square. It is in rectangular form .We having to find the squares on this rectangular board. So try to find out the 1 * 1 squares, 2 * 2 squares , 3 * 3 squares ....etc
Total number of 1 * 1 squares = 8 * 9
Similarly Total number of 2 * 2 squares = 7 * 8
Total number of 3 * 3 squares = 6 * 7
Total number of 4 * 4 squares = 5 * 6
...........
...........
Total number of 8 * 8 squares = 1 * 2
Total number of squares in an 8 * 9 board = (8 * 9) + (7 * 8) + (6 * 7) + (5 * 6) + .... + (1 * 2)
In general, total number of squares in an n * m board = n * m + (n-1) (m-1) + ... + 0

Or in other words, Total number of squares in a board which formed by join 'n' horizontal and 'm' vertical lines = (n-1) (m-1)

Total number of squares in 7 * 5 board?

  • A. 35
  • B. 325
  • C. 85
  • D. None of these

Solution:

Total number of squares = (7 * 5) + (6 * 4) + (5 * 3) + (4 * 2) + (3 * 1)

= 35+24+15+8+3

= 85

Ans: C

The micro-manometer in a certain factory can measure the pressure inside the gas chamber from 000001 to 999999 units. Lately this instrument has not been working properly. The problem with the instrument is that it always skips the digit '5' and move directly from 4 to 6. What is actual pressure inside the gas chamber if the meter displays 003016?

  • A. 2201
  • B. 2202
  • C. 2600
  • D. 2960

Solution:

The instrument doesn't report any number contains at least one '5'.

Find the number of numbers contains at least one '5'.

While considering the interval 1 to 100, there are the numbers contains at least one '5' are;

5, 15, 25, 35, 45, 50, 51, ..., 59, 65, 75, 85 and 95.

i.e. There are 19 numbers contains at least one '5'.

We can generalize this counting.

IntervalRequired number of numbers
1 to 100 10 + 10 - 1 = 19
101 to 200 19
201 to 300 19
301 to 400 19
401 to 500 20
501 to 600 99
601 to 700 19
701 to 800 19
801 to 900 19
901 to 1000 19
Total from 1 to 1000 271
Similarly;
1001 to 2000 271
2001 to 3000 271
3001 to 3016 2

Therefore the number of numbers contains at least one '5' from 1 to 3016 = 271 + 271 + 271 + 2

= 815

Hence, the actual reading when the meter shows 3016 = 3016 - 815

= 2201

Ans: A

Little Pika who is five and half years old has just learned addition. However, he does not know how to carry. For example, he can add 14 and 5, but he does not know how to add 14 and 7. How many pairs of consecutive natural numbers from 1000 and 2000(both included) can Little Pika add?

    • A. 150
    • B. 155
    • C. 156
    • D. 258
    • E. None of these

Solution:

Possible pairs should have the corresponding unit digits are;

(0 + 1), (1 + 2), (2 + 3), (3 + 4), (4 + 5) and (9 + 0).

Hence the possible consecutive pairs from the given range that he can add are given below.

1000 + 1001 1010 + 1011 ............. 1040 + 1041
1001 + 1002 1011 + 1012 ............. 1041 + 1042
1002 + 1003 1012 + 1013 ............. 1042 + 1043
1003 + 1004 1013 + 1014 ............. 1043 + 1044
1004 + 1005 1014 + 1015 ............. 1044 + 1045
1009 + 1010 1019 + 1020 ............. 1049 + 1050
Total 6 pairs Total 6 pairs   Total 6 pairs

In the above tabulation, there are 5 * 6 = 30 pairs are possible.

Similarly;

From 1100 to 1150 → 30 pairs

From 1200 to 1250 → 30 pairs

From 1300 to 1350 → 30 pairs

From 1400 to 1450 → 30 pairs

Therefore, in the above listed intervals put together, there are 5 * 30 = 150 pairs are possible.

In addition to the above listed possibilities, there are some other pairs are also possible.

1099 + 1100

1199 + 1200

1299 + 1300

1399 + 1400

1499 + 1500

1999 + 2000

Such 6 pairs are also possible.

Hence the total number of consecutive pairs = 150 + 6 = 156

Ans: C


A saint has a magic pot. He puts one gold ball of radius 1 mm daily inside it for ten days. If the weight of the first ball is 1 gm and if the radius of a ball inside the pot doubles every day, how much gold has the saint made due to his magic pot?

      • A)230 - 69/7
      • B)230 + 69/7
      • C)230 - 71/7
      • D)230 + 71/7
Solution:
First of all we have to catch the development of the weight of each ball from day one onwards.
About the ball puts in the first day:
Radius will increase in the sequence of the form; 1, 2, 4, 8, ....
i.e. The sequence of radius = 20, 21, 22,....29

i.e. the radius of the ball in the tenth day = 29 mm

Sequence of weight per day = 1, 8, 64, 512,.... = 80, 81, 82,..., 89

i.e. the weight of the ball in the tenth day = 89 gm

Hint:
If the radius = 1 mm → Volume = 4/3 π r3 = 4/3 π (1)3 = 1 (4/3 π) → Weight = 1 gm
If the radius = 2 mm → Volume = 8(4/3 π) → Weight = 8 gm
If the radius = 4 mm → Volume = 64(4/3 π) → Weight = 64 gm
And so on.
Similarly, the ball put in the second day will reach a weight of 88 gm in the tenth day and so on. i.e.
The ball put in the;Weight of the ball on the 10th day
1st day 89 gm
2nd day 88 gm
3rd day 87 gm
. .
9th day 81 gm
10th day 80 = 1 gm

 

Hence the total weight of all the balls on 10th day = 80 + 81 + 82 + ..... + 89

= (810 - 1)/(8-1) = (230 - 1)/7 gm

The initial weight of 10 balls = 10 gm

Therefore, the gain in weight = [(230 - 1)/7] - 10

= (230 - 1)/7 - 70/7

= (230 - 71)/7 gm

Ans: C

Fundamental rules of addition and multiplication

Rule 1: Addition principle

If there are 'n' independent events and the outcomes of each event is a, b, c, d..... respectively.

Then the total number of all the events = a + b + c + d +....

Example:

A box is containing 10 pens and 5 pencils. How many different ways you can select one pen OR one pencil from the box?

For selecting pen, there are 10 different ways.

For selecting pencil, there are 5 different ways.

For selecting one pen or one pencil, there are 10 + 5 = 15 ways possible.

In the above example the condition 'OR' has an important role, means 'OR' is dissolving the difference in identity of the product. i.e. either pen or pencil you have to select one product out of the total 15 products. Therefore there are 15 such possible ways.

Rule 2: Multiplication principle

If the number of events be 'm' and each of these 'm' events there are further 'n' events and which all are corresponding to the first set of 'm' events.

Then the number of ways to happen two events together is m * n.

Example:

A box is containing 10 pens and 5 pencils. How many different ways you can select one pen AND one pencil from the box?

For the selection of one pen, there are 10 different ways are possible.

Similarly for the selection of one pencil, there are 5 different possible ways.

Therefore the total number of possible ways to select one pen AND on pencil = 10 * 5 = 50 ways.

Example:

For reaching from City A to City B, there are 4 different roads and City C is connected with City B by 3 different roads. If a person needs to travel from City A to City C via City B, how many possible ways he can opt?

 

Number of ways from A to B = 4

Number of ways from B to C = 3

Total number of ways from A to C via B = 4 * 3 = 12

These 12 ways can be expressed in the form of ordered pairs, such as;

(a1, b1), (a1, b2), (a1, b3)

(a2, b1), (a2, b2), (a2, b3)

(a3, b1), (a3, b2), (a3, b3)

(a4, b1), (a4, b2), (a4, b3)

Result:
An event E1 can be completed in 'm' different ways and another event E2 can be completed in 'n' different ways, then the number of ways of the completion of;
  • E1 OR E2 is 'm + n'
  • E1 AND E2 is m * n'

 

Basic concept of arrangement (Repetition of objects is allowable)

This is fundamental practice of arrangements. Here we are using the fundamental principle of multiplication and addition.

Example: How many five digit numbers are multiple of 5?

In this type of questions, we can apply the concept of place arrangements.

Divisibility of 5 depends upon the unit digit of the number, i.e. unit digit should be either 0 or 5.

Ten thousands placeThousands placeHundreds placeTens placeUnit place
1, 2, ...9 0, 1, 2, ...., 9 0, 1, 2, ...., 9 Same as thousands place Same as ten thousands place
9 ways 10 ways 10 ways 1 way 1 way

Multiply all the possible ways to get the number of all the required numbers.

i.e. The number of required numbers = 9 * 10 * 10 * 10 * 2 = 18000

Conditional arrangements

This is the extension of the above concept. In this type we are considering some of the additional conditions in the arrangements, even though the repetition of the objects is allowable.

Example: How many 5 digits palindrome numbers are possible?

Palindrome numbers are numbers in which the extreme digits are same. While reading it from left to right or right to left, the value is same.

Ten thousands placeThousands placeHundreds placeTens placeUnit place
1, 2, ...9 0, 1, 2, ...., 9 0, 1, 2, ...., 9 Same as thousands place Same as ten thousands place
9 ways 10 ways 10 ways 1 way 1 way

Hence the total number of required numbers = 9 * 10 * 10 * 1 * 1 = 900

Example: How many 4 letter codes can be formed such that codes should start and end with a vowel and middle two letters are consonants?

There are 5 vowels and 21 consonants in the alphabet.

So the possible counting can be considered in the following manner.

1st place2nd Place3rd place4th Place
5 ways 21 ways 21 ways 5 ways

Therefore the total number of required codes = 5 * 21 * 21 * 5 = 11025

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