Code

First digit of code (odd) - 1,3,5,7,9 - 5 possibilities.

Last digit (even) - 0,2,4,6,8 - 5 possibilities.

5* _ _ _ _ * 5

Now we have 4 slots to fill with 8 numbers remaining, they can be filled in 8*7*6*5 ways.

In total, 5*(8*7*6*5)*5 = 42,000 ways. B IMO

I got differnet answer from B.. Please confirm my approach shankar (Thanks for your support)

P is for position:

P1: 5 ways
P2: 9
P3: 8
P4: 7
P5: 6
P6: 1 or 2 or 3 or 4 or 5

positions P2 till P5 can occupy 4 odd numbers or 4 even numbers
--if p2 to P5 has 4 odd numbers P6 has 5 even numbers to choose 1 out of it
--if p2 to p5 has 4 even number P6 has only 1 number to choose.

I got B

This is what I did

Even -- 0,2,4,6,8
Odd -- 1,3,5,7,9

First we will choose the 1st and last digit

1st -- 5 ways
last -- 5

so, out of 10, 2 numbers are out. Only 8 more left for selection
Rest can be selected in 8*7*6*5

Total --
5*5*8*7*6*5 = 42000

saketk wrote:I got B


First we will choose the 1st and last digit

42000

I totally agree with answer you got
but in your statement how to say that this is not an assumption "to choose first and the last initally". Else otherwise we are not getting an option as answer

Can you answer my above question to shankar??

Key entry logically starts with first to last ...

Well whenever you have conditions like these specified, try to consider them first as the others slots do not have any restrictions, so that we could directly do (8*7*6*5) as I did, in this case since repetition is not allowed we need to consider a lot of cases separately which would make it complicated (the way you're trying to solve)

So as a rule try to solve for given conditions (constraints) first and then you could solve for the remaining without any limitations.

Also note, any 6 digit integer without repetition can be formed in (9*9*8*7*6*5) ways = 136080.

You can clearly see options C,D and E are greater than this. We are definitely looking at an choice lesser than 136080. Hope it makes sense!!

HSPA wrote:I got differnet answer from B.. Please confirm my approach shankar (Thanks for your support)

P is for position:

P1: 5 ways
P2: 9
P3: 8
P4: 7
P5: 6
P6: 1 or 2 or 3 or 4 or 5

positions P2 till P5 can occupy 4 odd numbers or 4 even numbers
--if p2 to P5 has 4 odd numbers P6 has 5 even numbers to choose 1 out of it
--if p2 to p5 has 4 even number P6 has only 1 number to choose.

Your below point is awesome.. thanks...

shankar.ashwin wrote: Also note, any 6 digit integer without repetition can be formed in (9*9*8*7*6*5) ways = 136080.

You can clearly see options C,D and E are greater than this. We are definitely looking at an choice lesser than 136080. Hope it makes sense!!

HSPA wrote:I got differnet answer from B.. Please confirm my approach shankar (Thanks for your support)

P is for position:

P1: 5 ways
P2: 9
P3: 8
P4: 7
P5: 6
P6: 1 or 2 or 3 or 4 or 5

positions P2 till P5 can occupy 4 odd numbers or 4 even numbers
--if p2 to P5 has 4 odd numbers P6 has 5 even numbers to choose 1 out of it
--if p2 to p5 has 4 even number P6 has only 1 number to choose.

Also we just multiply numbers in permutation, so it really does not matter which position you consider first.

HSPA wrote:
I totally agree with answer you got
but in your statement how to say that this is not an assumption "to choose first and the last initally". Else otherwise we are not getting an option as answer

Can you answer my above question to shankar??

Key entry logically starts with first to last ...