In how many 5-digit numbers can be formed from the 10 digits (0-9), if
a) Repetitions are allowed ;
Two ways to get it:
First way:
The smallest is 10000 and the largest is 99999
There are 99999 integers from 1 to 99999 inclusive.
The integers from 1 to 9999 inclusive have fewer than 5 digits.
There are 9999 integers that have fewer than 5 digits.
99999-9999 = 90000 = the number of 5 digit integers.
Second way:
We can choose the 1st digit any of 9 ways, 1-9.
We can choose the 2nd digit any of 10 ways, 0-9.
We can choose the 3rd digit any of 10 ways, 0-9.
We can choose the 4th digit any of 10 ways, 0-9.
We can choose the 5th digit any of 10 ways, 0-9.
Answer 9*10*10*10*10 = 90000
b) The numbers begin with 40;
Two ways to get it:
First way:
The smallest is 40000 and the largest is 40999
There are 40999 integers from 1 to 40999 inclusive.
The integers from 1 to 39999 inclusive are less than 40000.
There are 39999 integers that are less than 40000.
40999-39999 = 10000 = the number that begin with 40.
Second way:
We can choose the 1st digit only 1 way, as 4.
We can choose the 2nd digit only 1 way, as 0.
We can choose the 3rd digit any of 10 ways, 0-9.
We can choose the 4th digit any of 10 ways, 0-9.
We can choose the 5th digit any of 10 ways, 0-9.
Answer 1*1*10*10*10 = 10000
c) The numbers are even;
Two ways to get it:
First way:
Half of the 90000 in the first part are even and half are odd.
Half of 90000 is 45000
Second way:
All even integers end in either 0,2,4,6 or 8.
We can choose the 1st digit any of 9 ways, 1-9.
We can choose the 2nd digit any of 10 ways, 0-9.
We can choose the 3rd digit any of 10 ways, 0-9.
We can choose the 4th digit any of 10 ways, 0-9.
We can choose the 5th digit any of 5 ways, 0,2,4,6, or 8.
Answer 9*10*10*10*5 = 45000
d) The numbers are divisible by 5.
Two ways to get it:
First way:
Every 5th integer is divisible by 5. Therefore:
One-fifth of the 90000 in the first part are divisible by 5.
One-fifth of 90000 is 18000
Second way.
All integers divisible by 5 end in either 0 or 5.
We can choose the 1st digit any of 9 ways, 1-9.
We can choose the 2nd digit any of 10 ways, 0-9.
We can choose the 3rd digit any of 10 ways, 0-9.
We can choose the 4th digit any of 10 ways, 0-9.
We can choose the 5th digit any of 2 ways, 0 or 5.
Answer 9*10*10*10*2 = 18000
ii) How many different sets of 4 students can be chosen out
of 17 qualified students to represent a school in a mathematics
contest?
17 students choose 4 = 17C4 = 1001.
Edwin
