The number of possible passwords under the restriction that letters are more than digits Good day, I am calculating this problem but I am not entirely sure if it is correct, so I would like to ask for some help.
I need to make a 4 character-long password from either digits or uppercase letters under the restriction that it has more letters than digits and that both can be repeated, and my initial thought is that there are two possible scenarios.

*

*3 letters plus 1 digit

*4 letters(no digits)

But I have no idea whether to use the combination formula or permutation formula as I stuck at how to pick 3 or 4 random positions from the password to insert letters, so I would like to ask for some help, thank you
 A: You have the right scenarios.
Now if there are $3$ letters and $1$ digit, first choose one of the $10$ digits and place it in one of the $4$ positions. Each of the rest three positions has choice of $26$ letters.
In case of all $4$ letters, each position in the password has $26$ choices.
Can you take it from here?
A: HINT:  If we denote a letter by L and a digit by D, then the possible patterns for passwords are:

*

*LLLD

*LLDL

*LDLL

*DLLL

*LLLL

Moreover, there are 26 possible letters and 10 possible digits.  Take it from there.
More generally, we can see that if we want an $n$-character password with exactly $d$ digits, there are exactly ${n \choose d}$ possible patterns of L's and D's for such a password.  One can calculate this for $0 \leq d \leq \lfloor n/2 \rfloor$, calculate the number of possible passwords for each such pattern, and then add up all the results.  I do not believe there is a more "elegant" answer to this question in general (though I will watch this thread to see if anyone comes up with one.)
