FUNDAMENTAL PRINCIPLES OF COUNTING

Problem 1 :

There are 10 boys and 8 girls. The teacher wants to select either a boy or a girl to represent in a function. In how many ways the teacher can make this selection ?

Solution :

Selecting a boy among 10 boys = 10

Selecting a girl among 8 girls = 8

Here we use fundamental principle of addition, because we select only one person either a boy or a girl. So,

= 10 + 8

= 18

Total number of ways is 18.

Problem 2 :

There are 3 candidates for a classical, 5 for Mathematical and 4 for Natural science scholarship.

i) In how many was can these scholarships be awarded ?

ii) In how many ways one of these scholarships be awarded ?

Solution :

i)  These three scholarships should be awarded, then

= 3 x 5 x 4 ways

= 60 ways

ii) Any one of these scholarships can be awarded :

= 3 + 5 + 4

= 12 ways.

Problem 3 :

A room has 6 doors. In how many ways can a man enter the room through one door and come out through a different door ?

Solution :

Number of ways to enter into room = 6

He has to come out to through a different room, so number of ways = 5

Both jobs should be done :

= 6 x 5

= 30 ways

Problem 4 :

The flag of a newly formed forum is in the form of three blocks, each to be colored differently. If there are six different colors on the whole to choose from, how many such designs are possible ?

Solution :

Number of colors available = 6

Required design will be in the form of 3 blocks.

3 options are available for first block, 2 options are available for 2nd block and 1 option is available for 1st block.

So, 

= 3 x 2 x 1

= 6

Problem 5 :

Find the number of 4 letter words with or without meaning which can be formed out of the letters of the word ROSE, when

i) The repetition of letters is not allowed 

ii) the repetition of letters is allowed

Solution :

i)  In the following blanks, we have to fill the letters

___   ___    ___   ___

so all 4 jobs should be done and we are allowed to use the letter once at a time. Then, we have to use fundamental principal of multiplication.

= 4 x 3 x 2 x 1

Here 4 will say, there are 4 letters available for 1^st place and so on.

= 24 ways

ii)  Repetition of letters is allowed :

= 4 x 4 x 4 x 4

= 256 ways

Problem 6 :

Given 4 flags of different colors, how many different signals can be generated if the signal requires the use of 2 flags one below the other ?

Solution :

Number of flags available = 4

Number of flags required to make the signal = 2

= 4 x 3

= 12 ways

Problem 7 :

Find the total number of ways of answering 5 objective type questions, each question having 4 choices.

Solution :

To answer 1^st question, we have 4 ways.

To answer 2^nd question, we have 4 ways and so on.

In this way, to answer 5 questions we have

= 4 x 4 x 4 x 4 x 4

= 4⁵ ways

Problem 8 :

There are 6 multiple choice questions in an examination. How many sequence of answers are possible, if the first three questions have 4 choices each and the next three have 5 each ?

Solution :

To answer first 3 questions, we have 

= 4 x 4 x 4

To answer last 3 questions, we have 

= 5 x 5 x 5

So, total number of ways are 

= 4 x 4 x 4 x 5 x 5 x 5

= 8000 ways

Problem 9 :

How many three digit numbers can be formed without using the digits 0, 2, 3, 4, 5 and 6 ?

Solution :

0, 1, 2, 3, ............. 9

We have 10 digits, from this the given digits are 0, 2, 3, 4, 5 and 6.

Number of digits available = 6

Excluding these 6 values, we have 4 options.

In 100^rds place, we can use any of the following digits 

1, 7, 8, 9

So, 4 options are available. Since repetition is allowed, we can use all four digits again.

So, the number of ways = 4 x 4 x 4

= 64 ways