## 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.

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