Thread: Password Combinations

Im slighty confused on how to calculate this.

Password must have length between 6 and 8.
Each password contains at least 1 digit [0-9].
Other characters either upper case or lower case.

So we have 62 different possiblities in each position, am i correct?

im thinking something like 62^6 + 62^7 + 62^8.
or 62! / 8! + 62! / 7! + 62! / 6!.

Please help me!

Hello, kurac!

Password must have length between 6 and 8.
Each password contains at least 1 digit [0-9].
Other characters either upper case or lower case.

So we have 62 different possiblities in each position, am i correct? . Yes!

Consider the 6-character passwords.

With no restrictions, there are: . possible passwords.

But the password must contain at least one digit.
We must eliminate the passwords with no digits (all letters).
. . There are: . passwords which have no digits.

Hence, there are: . six-character passwords
. . which contain at least one digit.


Similarly, there are:
. . seven-character passwords with at least one digit
. . eight-character passwords with at least one digit.

. * *

So the result of above is my answer?

Thanks.

Maths Helper, can you please also help me out with the question below?


How many binary strings of length 10 starts with 0 bit and ends with two 11 bits.


So, This is what ive done thus far.

0XXXXXXX11.

X can be either 1 or 0.

So, 10! / 7! 3! is what i have come up with.
But my problem is i havnt considered that each X can take 1 or 0.

So how to I do this?