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!.

2. 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!

With no restrictions, there are: .$\displaystyle 62^6$ possible passwords.

But the password must contain at least one digit.
We must eliminate the passwords with no digits (all letters).
. . There are: .$\displaystyle 52^6$ passwords which have no digits.

Hence, there are: .$\displaystyle 62^6 - 52^6$ six-character passwords
. . which contain at least one digit.

Similarly, there are:
. . $\displaystyle 62^7 - 52^7$ seven-character passwords with at least one digit
. . $\displaystyle 62^8 - 52^8$ eight-character passwords with at least one digit.

3. . * *

So the result of above is my answer?

Thanks.

4. 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?