Problem Solving

38. If 6 different numbers are to be selected from integers 0 to 6, how many 6-digit even integers greater than 300,000 can be composed?

OA is 1680 ?????

imo 2400

28812?

Basically there are 7 digits (0,1,2,3,4,5,6). So,
3 * 7 * 7 * 7 * 7 * 4.

4 because it has to be even and 3 because the digits have to be greater than 3,00,.

Please confirm.

enniguy wrote:28812?

Basically there are 7 digits (0,1,2,3,4,5,6). So,
3 * 7 * 7 * 7 * 7 * 4.

4 because it has to be even and 3 because the digits have to be greater than 3,00,.

Please confirm.

Wouldn't you want to included 3....300,002 > 300,000?

I think it is 2880.

4*6*5*4*3*2*1=2880 let me know if it is correct?

I calculated as follows:

7 usable digits (0,1,2,3,4,5,6)

We need to find numbers greater than 3,00,000. hence,
1) [3] [] [] [] [] [last digit can be any of 0,2,4,6.ie. 4 ways to select last digit]
2) [4] [] [] [] [] [any of 4 even digits]
3) [5] [] [] [] [] [any of 4 even digits]
4) [6] [] [] [] [] [any of 4 even digits]
We have used up 2 out of 7 digit, we are left with 5 digits
Total number of ways to select 1st selection = 1*5*4*3*2*4 = 480 ways
but this is similar to remaining 3 cases...hence each case will have 480 ways

total number of ways = 480*4 = 1920 ways.

Is this correct? if not what am I missing and where?

Almost correct...

But,

We need to find numbers greater than 3,00,000. hence,
1) [3] [] [] [] [] [last digit can be any of 0,2,4,6.ie. 4 ways to select last digit]
2) [4] [] [] [] [] [any of 3 even digits] (already chosen one even digit)
3) [5] [] [] [] [] [any of 4 even digits]
4) [6] [] [] [] [] [any of 3 even digits]


Hence
for 1) we have 480 combinations- 5.4.3.2.4
for 2) we have 360 combinations- 5.4.3.2.3
..
..


so in all 480*2+360*2=1680