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 = 8^{2 }

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

Similarly, total number of 2 * 2 squares = 7 * 7 = 7^{2}

Total number of 3 * 3 squares = 6 * 6 = 6^{2}

.

.

Total number of 8 * 8 squares = 1 * 1 = 1^{2}

Total number of squares on a chess board = 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} + 7^{2} + 8^{2}

= (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 = 1 Total number of squares in a 9 * 9 square board = 1 Total number of squares in an N * N square board = 1 Or in other words, total number of squares in a board which is formed by intersecting 'n' horizontal and 'n' vertical lines = 1 |

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

Solution:

Number of squares = 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} + 7^{2} + 8^{2} + 9^{2} + 10^{2}

= 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} = 6^{2} = 36

Total number of 4 * 4 squares on an chess board = (8 - 4 + 1)^{2} = 5^{2} = 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} = 17^{2} = 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} = 41^{2} = 1681

Ans: C

Total number of squares in an m * n boardExample: 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