how many 4 digit number.....

It's hard not to make errors, need to think to everything all the time.


"ending with 2
02 12 32 52 62
it would be again 4*3*5 = 60 "

3*3 for 12 32 52 62 due to the 0 which cannot be on first place

Same for:
"ending with 6
06 16 26 36 56 again 60"

Sudhir you are doing everything right except you are mixing the values of 02, 12,32,52,62

Your first step is absolutely right. 4*3*5 = 60

In the next step try isolating the 0 values i.e. calculate the combination of
02 and 06

and then calculate for 12,32,52,62

and then 16,26,36,56

The reason we are isolating the values of 02 and 06 is because if we mix them together our equation of permutation becomes inconsistent.

e.g. 02,32,52,62 we know that we cannot put 2 in the place of first digit because it already used and we also cannot put 1 more digit in the first place since it used, but we cannot put 0 in the first place as well. this becomes inconsistent with everything.

I am sure you will find the answer with the above method, if you are unable to just let me know i'll guide you through.

i got my mistake..i shudnt include 02 and 06 in the list of ending with 2' and 6.. as these are special case

02 we have 4*3*1 = 12

similarly with 06 = 12

ending with 2 will have 12 32 52 62 hence 3*3*4= 36 ways..

similarly with 6 we have 16 26 36 56 hence 3*3*4

therefore its 60+36+36+12+12
156 ..wow..cracked it,,,


thanksa lot "pepeprepa" and parallel chase....