FUNDAMENTAL PRINCIPLE OF COUNTING PROBLEMS WITH SOLUTION

Problem 1 :

A person went to a restaurant for dinner. In the menu card, the person saw 10 Indian and 7 Chinese food items. In how many ways the person can select either an Indian or a Chinese food?

Solution :

Number of ways of selecting Chinese food items  =  7

Number of ways of selecting Indian food items  =  10

Here a person may choose any one food items, either an Indian or a Chinese food. So, we have to use "Addition" to find the total number of ways for selecting the food item.

Total number of selecting Indian or a Chinese food 

=  10 + 7  =  17 ways

Problem 2 :

There are 3 types of toy car and 2 types of toy train available in a shop. Find the number of ways a baby can buy a toy car and a toy train?

Solution :

According to the given question, a baby wants to buy both toy car and toy train. So we have to use the binary operation "Multiplication" to find the total number of ways.

Types of car available in the shop  =  3

Types of car available in the shop  =  2

Total number of ways of buying a car  =  3 (2)  =  6

Problem 3 :

How many two-digit numbers can be formed using 1,2,3,4,5 without repetition of digits?

Solution :

In order to form a two digit number, we have to select two numbers out of the given 5 numbers. 

___  ___

To fill up the first dash, we have 5 options.

To fill up the second dash, we have 4 options.

Number of two digit numbers formed using the above numbers  =  5 (4)  =  20.

Problem 4 :

Three persons enter in to a conference hall in which there are 10 seats. In how many ways they can take their seats?

Solution :

1^st person may choose 1 seat out of 10 seats

2^nd person may choose 1 seat out of 9 seats

3^rd person may choose 1 seat out of 8 seats

Total number of ways of selecting seat  =  10 (9) (8)

  =  720 ways

Problem 5 :

In how many ways 5 persons can be seated in a row?

Solution :

5 persons may sit in 5 seats.

1st person may sit any one of the 5 seats 

2nd person may sit any one of the 4 seats and so on.

Hence the total number of ways  =  5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1

 =  120 ways

Problem 6 :

If each of the students of in a class of 30 students is capable of winning any of the class prizes, how many ways are there for awarding ?

a) a first prize, a second prize, and a third prize in mathematics ?

b) a mathematics prize, a chemistry prize, a physics prize ?

Solution :

Total number of students in the class = 30

a) In those 30 students each person is eligible to be awarded. 

In this way, 29 students are eligible to be awarded second prize. 28 students are eligible to be awarded the third prize.

Total number of options = 30 x 29 x 28

= 24360

b) Here the prizes are given based on the subjects. All 30 students are eligible to awarded to get prizes in the subjects math, physics and chemistry.

= 30 x 30 x 30

= 27000

Problem 7 :

Three digits numbers are formed using only the digits

2, 3, 5, 6, 7 and 9

a) if the repetition are not permitted, how many 3 digit numbers can be formed ?

b)  how many of these are 

i) less than 400    ii)  even     iii)  odd    iv) multiples of 5

Solution :

The digits are 2, 3, 5, 6, 7 and 9

Total options = 6

a) We create 3 digit number and repetitions are not allowed

• number of options available to fill the unit digit = 6
• number of options available to fill the tens digit = 5
• number of options available to fill the hundreds digit = 4

Total number of options = 6 x 5 x 4

= 120

b)

i) When the number is lesser than 400, it should start with 2 or 3. 

• Number of options to be filled in the hundred's digit = 2
• Number of options to be filled in the ten's place = 5
• Number of options to be filled in the unit place = 4

= 2 x 5 x 4

= 40

ii)  even

To have even number, we should have any one of the digits 2 or 6.

Available options to fill the one's place = 2

Available options to fill hundred's place = 5

Available options to fill hundred's  place = 4

Total number of options = 4 x 5 x 2

= 40

iii)  odd

To have odd number, we should have any one of the digits 3, 5, 7 or 9.

Available options to fill the one's place = 4

Available options to fill hundred's place = 5

Available options to fill hundred's  place = 4

Total number of options = 4 x 5 x 4

= 80

iv) multiples of 5

To have a multiple of 5, the ending digit should be 5.

Number of ways = 4 x 5 x 1

= 20 ways

Problem 8 :

A vehicle license plate consists of 3 letters followed by 3 digits. How many different license plates are possible if 

a)  there are no restrictions on the letters or digits used ?

b)  no letters may be repeated 

c) The first digit cannot be 0 and no digits can be repeated ?

Solution :

a)  Total number of digits = 10 (o to 9)

Total number of alphabets = 26

First three should be letter and next three should be numbers.

= 26 x 26 x 26 x 10 x 10 x 10

= 17576000

b) No letters may be repeated

= 26 x 25 x 24 x 10 x 10 x 10

= 15600000

c) 

= 26 x 26 x 26 x 9 x 9 x 9

= 11389248

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