Question

Consider four-digit numbers that consist of 0, 1, 2, 5, 6, and 9.
a) How many four-digit numbers can be formed from the digits 0, 1, 2, 5, 6, and 9 if each digit can be used
only once? (the four-digit numbers can't start with 0).
b) How many of those four-digit numbers are even?

c) How many are greater than 2200?

Answer 1

Four digit numbers using the digits 0, 1, 2, 5, 6 and 9
(a)
To form a four digit number (each digit being used only once) we have to select 4 unique digits from (0,1,2,5,6,9).
So since first digit can't be 0. So select 1st digit in 5 ways (1,2,5,6,9)
Select 2nd digit in 5 ways.
Select 3rd digit in 4 ways
Select 4th digit in 3 ways
Total number of 4 digit numbers with unique digits = 5 x 5 x 4 x 3 = 300
(b)
Since the number has to be even, the 4th digit should be either of (0,2,6)
So we divide the problem in two cases
Case 1: 4th digit is 0.
So select 1st digit in 5 ways (1,2,5,6,9)
Select 2nd digit in 4 ways.
Select 3rd digit in 3 ways
4th digit is 0.
Total number of 4-digit numbers with 0 as the 4th digit = 5 x 4 x 3 = 60
Case 2: 4th digit is non-zero
Select 4th digit in 2 ways ( 2, 6) (Number should be even)
Select 1st digit in 4 ways (Can't use 0)
Select 2nd digit in 4 ways
Select 3rd digit in 3 ways
Total = 2 x 4 x 4 x 3 = 96
Total even numbers of 4 unique digits = 60 + 96 = 156
(c)
For a number greater than 2200, 1st digit should either be 2 or greater than 2.
Case 1: 1st digit is 2.
Now second digit should be either 5, 6 or 9, since 2 has been used (each digit can be used only once) and the number has to be greater than 2200.
Select 2nd digit in 3 ways (5,6,9)
Select 3rd digit in 4 ways
Select 4th digit in 3 ways
Total = 3 x 4 x 3 = 36
Case 2: 1st digit is either one of 5,6 and 9
Select 1st digit in 3 ways (5,6,9)
Select 2nd digit in 5 ways
Select 3rd digit in 4 ways
Select 4th digit in 3 ways
Total = 3 x 5 x 4 x 3 = 180
Total numbers that are greater than 2200 = 36 + 180 = 216