FUNDAMENTAL PRINCIPAL OF COUNTING MULTIPLICATION RULE

If there are two jobs such that one of them can be completed in m ways, and when it has been completed in any one of these "m" ways, second job can be completed in "n" ways; then the two jobs in succession can be completed in

 m × n ways.

Example 1 :

Twelve students compete in a race. In how many ways first three prizes are given?

Solution :

We have to select 3 students for distributing first three prizes.

Total number of students participating in the race  =  12

All 12 students are having equal chance to get the 1^st prize.

The number of ways of selecting a student who got second prize out of 11 students  =  11

The number o f ways of selecting a student who got third prize out of 10 students  =  10

The number of ways for distributing first three prizes

=  12 x 11 x 10

=  1320

Example 2 :

From among the 36 teachers in a college, one principal, one vice principal and the teacher in-charge are to be appointed. In how many ways this can be done?

Solution :

Total number of teachers  =  36

The number of ways of selecting one principal out of 36 teachers  =  36

The number of ways of selecting one vice principal out of remaining 35 teachers  =  35

The number of ways of selecting one teacher in-charge out of remaining 34 teachers = 34

The number of ways of selecting one principal, one vice principal

=  36 x 35 x 34

=  42840

Example 3 :

There are 6 multiple choice questions In an examination, How many sequence of answers are possible, if the first three have 4 choices each and next three have 2 each?

Solution :

Number of chances for first three questions = 4

Number of possible ways of getting correct answer = 4

Number of chances for next three questions = 2

Number of possible ways of getting correct answer = 2

Number of ways of getting correct answer for first three questions

=  4 x 4 x 4

=  64

Number of ways of getting correct answer for next three questions

=  2 x 2 x 2

=  8

Total number of ways  =  64 x 8

=  512

Example 4 :

How many 3-letter code words can be formed if at least one of the letters is to be chosen from the vowels a, e, i, o, and u

Solution :

Given letters are a, e, i, o and u

Total number of distinct letters = 5

Number of ways to choose first letter = 5

Number of ways to choose second letter = 4

Number of ways to choose third letter = 3

Total number of ways = 5 x 4 x 3

= 60

Example 4 :

There are 5 questions each have four options. Then in how many different ways can be answer the question ?

Solution :

Number of options to answer the question = 4

Number of questions = 5

Number of ways = 4 x 4 x 4 x 4 x 4

= 1024

So, 1024 ways are there to answer 5 questions.

Example 5 :

Suppose you have totally forgotten the combination to your locker. There are three numbers in the combination, and you’re sure each number is different. The numbers on the lock’s dial range from 0 to 35. If you test one combination every 12 seconds, how long (in days to the nearest hundredth) will it take to test all possible combinations?

Solution :

Number of 1 digit numbers from 0 to 9 = 10

Number of 2 digit numbers

from 10 to 19 = 10 numbers 

from 20 to 29 = 10 numbers

from 30 to 35 = 6 numbers 

= 10 + 10 + 6

= 26 numbers

10 + 26 

= 36 numbers

Here all 36 numbers are having equal chances to get selected.

Number of ways = 36 x 35 x 34

= 42840

Number of times she is doing combination = 42840 x 12

= 514080

Converting to days = 514080/(60 x 60 x 24)

= 5.95 days

Example 6 :

How many license plates of 3 symbols (letters and digits) can be made using at least 2 letters for each?

Solution :

First case: all 3 symbols are letters:

26 x 26 x 26 = 17,576 possible ----(1)

Second case: 2 letters & one digit:

10(26 x 26) = 6,760

by considering the one digit to be qt extreme right.

But there are two other possibilities--one digit in middle & one digit at extreme left.

There this accounts for 3(6760) = 20,280----(2)

possible "second case" arrangements

(1) + (2)

= 17,576 + 20280

= 37856 ways

Example 7 :

How many positive odd integers less than 10,000 can be written using the digits 3, 4, 6, 8, and 0?

Solution :

The number which is lesser than 10000 will be 1 digit number, 2 digit number, 3 digit number and 4 digit number.

Number of 1 digit number :

1 number

Number of 2 digit number :

43, 63, 83, 33 ==> 4 numbers

Number of 3 digit number :

• Number of options is available on one's place = 1
• Number of options available on 10's place = 5
• Number of options available on 100's place = 4

4 x 5 x 1 => 20

Number of 4 digit number :

• Number of options is available on one's place = 1
• Number of options available on 10's place = 5
• Number of options available on 100's place = 5
• Number of options available on 1000's place = 4

4 x 5 x 5 x 1 => 100

Total number of ways = 100 + 20 + 4 + 1

= 125

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