How many nos greater than 3000 can be formed from 1,2,3,4,5 without repeating any of them?How many willbe even

Question

How many of them will also be odd? Please include workings and explanations. thanks

Answer 1

3*4*3*2= 72

1st digit = 3,4,5

How many will be even
=> 2*3*2*2 = 24.........1st digit = 3,5 & last = 2,4
=> 1*3*2*1= 6......1st digit = 4 & last = 2
add ans = 30

How many of them will also be odd?
30 even & rest are odd

Answer 2

The 1st digit can be 3 or 4 or 5 (3 possibilities)
The 2nd digit can be any of the remaining 4 digits
the 3rd digit can be any of the remaining3 digitsfinal digit can be either of the remaining 2 digits, so total possibilities are 3*4*3*2 = 72.
The even numbers will be those ending in 2 or 4
If the number starts with a 3 or 5 there will be 16 ways to get a number ending in 2 or 4. If the number begins with 4 then there will be 6 ways to a number ending in 2. Thus tere will be 24 even numbers.

Answer 3

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.

Hope this helped.

Regards, :)

Answer 4

Total digits given are 5. Number is > 3000 so you have to keep either 3, 4 or 5 in 1000's place. ie. you can keep 3 digits in 1000's place.
Then you can keep remaining 4 digits in 100's place
Then you can keep remaining 3 digits in 10's place
Then you can keep remaining 2 digits in unit's place
Total numbers formed = 3 × 4 × 3 × 2 = 72

To form even numbers,
you have to put 2 or 4 in unit's place. these two digits can be put in unit's place.
Then you can put either 3 or 5 in 1000's place. 2 digits in 1000's place.
So digits remained for 100's place = 3 and for 10's place 2
No. formed = 2 × 3 × 2 × 2 = 24

For odd numbes, you can put 3, 4 or 5 in 1000's place,
digits 1, 3 or 5 in units place.
Digits remained for 100's place = 3 and for 10's place = 2
No. formed = 3 × 2 × 1 × 3 = 18