@SVIX: According to my calculation, this gives 2,992,430,052,218,880,000, almost 12 times the correct answer, 251,471,033,958,144,000, that I gave earlier in posting http://groups.google.com/group/algogeeks/msg/bb2269736a997419. This is because you are counting some passwords multiple times. Consider, for example, the set of passwords that have 3 "A"s, 3 "a"s, 2 "1"s, and 2 additional "A"s for the "at large" characters. You would say that there are 10C3 * 7C3 * 4C2 = 25,200 of them. The mistake you are making is distinguishing between the first 3 "A"s and the last 2, whereas they actually are indistinguishable. In actuality, there are 10C5 * 5C3 * 2C2 = 2,520 of them.
Dave On Feb 9, 10:38 pm, SVIX <[email protected]> wrote: > 1. there should be atleast 3 capital letters > 2. atleast 3 small letters > > -> 6 spaces gone for these, with repetitions allowed. > > for 3 spaces, we have 26^3 possibilities, and they can be arranged in > 10C3 ways... > > for the next 3, they can be arranged in 7C3 ways > > 3. atleast 2 numbers 0-9 > > now, 4 spaces left, 10 digits, 100 combinations.. 4C2 ways > > 4. the password should has length=10 > > remaining 2 spaces. assuming only caps, small and numbers, we have > (26+26+10)^2 combinations. > > On Feb 9, 5:46 am, 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- Hide quoted text - > > - Show quoted text - -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
