How many 3-digit numbers can be formed by choosing from 1 2 3 and 4 if repetition is not allowed

India's Super Teachers for all govt. exams Under One Roof

Enroll For Free Now

Concept:

Fundamental Principle of Multiplication:

Let us suppose there are two tasks A and B such that the task A can be done in m different ways following which the second task B can be done in n different ways. Then the number of ways to complete the task A and B in succession respectively is given by: m × n ways.

Fundamental Principle of Addition:

Let us suppose there are two tasks A and B such that the task A can be done in m different ways and task B can be completed in n ways. Then the number of ways to complete either of the two tasks is given by: (m + n) ways.

Calculation:

Here, we have to find how many 3-digit numbers can be formed without using the digits 0, 2, 3, 4, 5 and 6.

i.e we have to find how many 3-digit numbers can be formed using the digits 1, 7, 8, 9.

Clearly, repetition of digits is allowed.

The number of ways to fill unit's digit = 4

The number of ways to fill ten's digit = 4

The number of ways to fill hundred's digit = 4

∴ Total number of required numbers = 4 × 4 × 4 = 64

India’s #1 Learning Platform

Start Complete Exam Preparation

Daily Live MasterClasses

Practice Question Bank

Mock Tests & Quizzes

Get Started for Free Download App

Trusted by 3.4 Crore+ Students

Text Solution

Solution : (i) When repetition of digits is allowed:
No. of ways of choosing firsy digits = 5
No. of ways of choosing second digit = 5
No. of ways of choosing third digit = 5
Therefore, total possible numbers `= 5 xx 5 xx 5 = 125`
(ii) When repetition of digits is not allowed:
No. of ways of choosing first digit = 5
No. of ways of choosing second digit = 4
No. of ways of choosing thrid digit = 3
Total possible numbers `= 5 xx 4 xx 3 = 60`.