Question

How many +ve numbers can be formed by using any number of the digits 0 , 1 , 2 , 3 , and 4 with no digit being repeated?

Answer 1

★ COMBINATORICS ★

Number of digits , n = 5
Number of numbers of 1 digit = 4

To find the number of numbers of 2 digits number of places to be filled up r = 2

Number of digits , n = 5
Number of numbers of 2 digits = 5!/3! = 20

Out of these 20 numbers , some numbers begin with 0 ,
The Number of such numbers of digits beginning with 0 , number of places and number of digits remains to be utilized aslike , n = 4

Hence , number of numbers of 2 digits which begin with 0 = 4 ,
Number of numbers of 2 digits = 20 - 4 = 16

Similarly , number of numbers with 3 digits = 5!/2! = 60

Again rejecting non utility cases :

Number of numbers of 3 digits beginning with 0 is 12 ,

Number of numbers of 3 digits = 60 - 12 = 48

Similarly :

Number of numbers of 4 digits = 5! = 120

After resolution :

Number of numbers of 4 digits = 120 - 24 = 96

Number of numbers of 5 digits = 120

Number of numbers of 5 digits which begin with 0 = 4! = 24

HENCE , Number of numbers of 5 digits = 120 - 24 = 96

Number of +ve numbers formed from 0 , 1 , 2 , 3 , 4 , without repetition is :
4 + 16 + 48 + 96 + 96 = 260

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