No. That is too large, at 2,087,438,895,360,000,000. Analyzing your
expression, 26C3 is the number of ways to choose 3 different letters,
but the letters can be the same. The number of combinations of 3
letters with repetitions is 26^3. So that aspect of your formula is
too small. However, you can't multiply by 10! because when there are
duplicate letters or digits, you will be counting those duplicates
multiple times. E.g., under your scheme, passwords with 3 "A"s, 3
"a"s, and 4 "1"s will be counted 10! times when there are only 10! /
(3! * 3! * 4!) = 4,200 distinct combinations.

My expression says, e.g., that there are 10C3 different combinations
of positions within the 10 character password where you can put 3
capital letters and there are 26^3 combinations of the letters. Then,
once you've placed 3 capital letters, there are 7 spaces left, and you
can put the 3 lower case letters in 7C3 different combinations of
those positions, again with 26^3 combinations of letters, and finally,
the four remaining positions contain digits, and there are 10^4
posibilities. Etc.

Dave

On Feb 9, 1:48 pm, Manmeet Singh <[email protected]> wrote:
> Also it shud now be multiplied with Factorial of 10
>
> On Thu, Feb 10, 2011 at 1:14 AM, Manmeet Singh <[email protected]>wrote:
>
>
>
> > @ Dave : I think this shud also give the same result
> > C(26, 3) * C(26, 3) * C(10, 2) *  C(62, 2)
>
> > On Wed, Feb 9, 2011 at 7:16 PM, snehal jain <[email protected]> wrote:
>
> >> how many passwords can be made if
> >> 1. there should be atleast 3 capital letters
> >> 2. atleast 3 small letters
> >> 3. atleast 2 numbers 0-9
> >> 4 the password should has length=10
>
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