FUNDAMENTAL PRINCIPLE OF COUNTING PROBLEMS

Problem 1 :

A mobile phone has a passcode of 6 distinct digits. What is the maximum number of attempts one makes to retrieve the passcode?

Solution :

The passcode must be formed using the following digits

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Each digits must be different.

Total number of ways  =  10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5

  =  151,200

Since we have to use distinct digits, we cannot choose the repeated numbers.

Hence the total number of ways of retrieving the passcode is 151200.

Problem 2 :

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

Solution :

1^st flag may be chosen out of 4 flags, 2^nd flag may be chosen out of 3 flags and 3^rd flag may be chosen out of 2 flags.

Total number of ways  =  4 ⋅ 3 ⋅ 2

  =  24 ways

Hence 24 different signals may be formed using 4 flags.

Problem 3 :

Four children are running a race.

(i) In how many ways can the first two places be filled?

(ii) In how many different ways could they finish the race?

Solution :

(i) Out of 4 children, any one may get first prize. Out of three children, any one may get the second prize.

Hence the total number of ways  =  4 (3)  =  12 ways.

(ii)  Out of 4 ----> Any one may get 1^st prize

Out of 3 ----> Any one may get 2^nd prize

Out of 2 ----> Any one may get 3^rd prize

Remaining 1 will get fourth place.

Hence total number of ways  =  4 ⋅ 3 ⋅ 2 ⋅ 1  =  24 ways

Problem 4 :

Count the number of three-digit numbers which can be formed from the digits 2, 4, 6, 8 if (i) repetitions of digits is allowed. (ii) repetitions of digits is not allowed

Solution :

Required three digit number

  ___   ___   ___

(i) repetitions of digits is allowed

Since repetition of digits is allowed, we have 4 options to fill each places.

Hence the numbers to be formed with the given digits are 

  =  4 ⋅ 4 ⋅ 4 

  =   64

(ii) repetitions of digits is not allowed

Hundred place :

We may use any of the digits (2, 4, 6, 8), so we have 4 options.

Tens place :

Repetition is not allowed. So, we have 3 options.

Unit place :

By excluding the number used in the hundreds and tens place, we have 2 options.

Hence total numbers to be formed  =  4 ⋅ 3 ⋅ 2

  =  24

Problem 5 :

Determine how many different computer passwords are possible if:

a) digits and letters can be repeated, and

(b) digits and letters cannot be repeated.

(i) 3 digits followed by 4 letters

(ii) 2 digits followed by 5 letters.

Solution :

Digits are 0, 1, 2, 3, 4, ...............9

Total number of digits = 10

Letters are a, b, c, ...............z 

Total number of letters = 26

(i) 3 digits followed by 4 letters

When digits are repeating and letters are repeating,

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

= 456976000

When digits and letters are not repeating.

= 10 x 9 x 8 x 26 x 25 x 24 x 23

= 258336000

(ii)  2 digits followed by 5 letters.

When digits are repeating and letters are repeating,

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

= 1188137600

When digits and letters are not repeating.

= 10 x 9 x 26 x 25 x 24 x 23 x 22

= 710424000

Problem 6 :

A men’s department store sells 3 different suit jackets, 6 different shirts, 8 different ties, and 4 different pairs of pants. How many different suits consisting of a jacket, shirt, tie, and pants are possible?

Solution :

Number of suit jackets = 3

Number of shirts = 6

Number of ties = 8

Number of pants = 4

Possible number of ways = 3 x 6 x 8 x 4

= 576

So, total number of possible ways are 576.

Problem 7 :

A baseball manager is determining the batting order for the team. The team has 9 players, but the manager definitely wants the pitcher to bat last. How many batting orders are possible?

Solution :

Total number of players = 9

Since the pitcher should bat last, there is only one choice. By leaving the pitcher, we have only one option.

= 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 x 1

= 40320

Problem 8 :

How many eight-digit numbers can be formed if the leading digit cannot be a zero and the last number cannot be 1?

Solution :

Total number of digits = 10 (including 0 and 9)

To fill the first digit, we have 9 options excluding 0

To fill the last digit, we have 9 options excluding 1.

= 9 x 10 x 10 x 10 x 10 x 10 x 10 x 9

= 81000000

Problem 9 :

How many 4 digit odd numbers can be formed if no digit can be repeated?

Solution :

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

Even numbers = 0, 2, 4, 8

Odd numbers = 1, 3, 5, 7, 9

Number which ends with 0, 2, 4, 6, 8 is even number.

To fill the last digit, we have 5 options

To fill the remaining places, we have excluding 0 and the number what we fill in the last place, we have 8 options.

= 8 x 8 x 7 x 5

= 2240

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