A Number Theory Problem: “How many numbers greater than 3000 can be formed from 1,2,3,4,5 without repeating any of them? How many will be even?”

Answer

There are two type of numbers greater than 3000 that can be formed here namely, 4 digit numbers and 5 digit numbers,.

Note that no repetition allowed.

For the 4 digit numbers,
First digit can be filled in ‘3’ ways ( from either 3,4, or 5 )
Second digit can be filled in ‘4’ways ( any of the remaining 4 numbers after using one for the first place. )
Third digit can be filled ‘3’ ways ( any of the remaining 3 numbers after usage above. )
Fourth digit can be filled ‘2’ ways ( any of the remaining 2 numbers after usage above . )

Thus here there are 3*4*3*2 = 72 numbers that are greater than 3000 but 4 digits in nature in our problem.

For the 5 digit numbers,
First digit can be filled in ‘5’ ways ( any of the numbers can fill this position )
Second digit can be filled in ‘4’ ways ( any of the remaining numbers can fill this position )
Third digit can be filled in ‘3’ ways ( any of the remaining numbers can fill this position )
Fourth digit can be filled in ‘2’ ways ( any of the remaining numbers can fill this position )
Fifth digit can be filled in ‘1’ ways ( since only one number will remain after all used up ‘above’ )

Thus here there are 5*4*3*2 = 120 numbers that are greater than 3000 but 5 digits in nature in our problem.

Giving a total of 72 + 120 = 192 numbers for the first part of the problem.

How many are even?

For the 4 digit numbers, ( even case )
if the number starts at 4, we have last place must be ‘2’. The remaining numbers can be ordered in two places in 3*2 = 6 ways giving 6 numbers for this case.
if the number starts at 3 or 5, there are ‘2’ ways for the first digit and ‘2’ ways for the last digit ( either 2 or 4 to ensure even); Hence giving a the No. of 2*2*3*2 = 24 numbers in this case.

A total of 6 + 24 = 30 even 4 digit numbers. Since total 4 digit numbers was 72, thus total odd numbers would be 72 – 30 = 42. That is 42 four digit odd numbers that satisfy our problem.

For the 5 digit numbers ( even case ),
The last digit can be filled in ‘2’ ways ( either 2 or digit 4 ). The rest of the digits may be permutated in any way. Hence, the No. in this case is 2*4*3*2*1 = 48. That is we have 48 even 5 digit numbers to our problem. Since total 5 digit numbers that satisfy our problem is 120, this means that we have a total of 120 – 48 = 72 five digit odd numbers that satisfy our problem.

To Summarize,the answer to our problem is :

72 Four digit numbers consisting of 30 even and 42 odd.

120 Five digit numbers consisting of 48 even and 72 odd.

192 numbers in total that satisfy our problem.