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".
- 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 numbers | Number of numbers | Required key-depressions for each number | Total 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.
| Interval | Required 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
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. |
| 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;
|
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 place | Thousands place | Hundreds place | Tens place | Unit 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 place | Thousands place | Hundreds place | Tens place | Unit 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 place | 2nd Place | 3rd place | 4th Place |
|---|---|---|---|
| 5 ways | 21 ways | 21 ways | 5 ways |
Therefore the total number of required codes = 5 * 21 * 21 * 5 = 11025
