Problem Solving

with Numbers ending in 2
last 2 digit combinations could be 12,32,52,72.
So we have 5*5*4 such nos

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In that case you need a 4 digit number so you must not have 0 on the first digit or it is 0342 for example, it is not a 4 digit number.

For example _ _ 1 2
1st digit: you choose among 3 4 5 6 7 = 5 choices
2nd digit you choose among 0 3 4 5 6 7 - the digit you chose for the 1st digit= 5 choices

And there are 4 combinations of last 2 digits (12,32,52,72)
So 5*5*4

2nd digit you choose among 0 3 4 5 6 7 - the digit you chose for the 1st digit= 5 choices

i thought it shud be 0 3 4 6 7 ( assuming the number 5 has been chosen in as the first digit ) because no digit can occur more than once.. correct if i am wrong..

Thanks a lot all for clearing my doubts ,,, i know i am pushing it a bit too long.... but want to make sure my basics are right.

When I say "2nd digit you choose among 0 3 4 5 6 7 - the digit you chose for the 1st digit= 5 choices"
I say there are 5 choices which are possible, not I choose 5.
I think it's why you ask.

thanks a lot for all ur help guys...

Finally i can say i know how to solve the question... thanks a lot...

hey just for my understanding i am posting another question.. please let me know if its right...

how many 4 digit number that are divisble by 2 can be formed using digits (0 1 2 3 5 6 ) if no digit is to occur more than once in each number

for a number to divisble by 2 . it should end with 0, or any multiple of 2.( even number)


ending with 0 would be

10 20 30 50 60

hence its wud be 4*3*5= 60


ending with 2

02 12 32 52 62

it would be again 4*3*5 = 60


ending with 6

06 16 26 36 56 again 60

hence it shud be 180 .. but the answer is 156..i guess he is taking the ending with 2 and ending with 6 numbers as 4*3*4= 48

hence 60+48+48=156..

please let me know where i am going wrong on this one...

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....