A question about "ways" in Math

Question

How many ways are there to make a two-character alphanumeric password, but it must include an integer?

The solution is: $36^2-26^2$.

However, why can't it be: $36×10$? Since one 'slot' is a forfeit for the integer.

Answer 1

Your problem is that by using $2(36)(10)$, you are counting each digit-digit combination twice, so you get $200$ digit-digit combinations.

Keeping that in mind, notice the following:

• There are $10^2$ digit-digit combinations.
• There are $26(10)$ letter-digit combinations.
• There are $26(10)$ digit-letter combinations.

Hence, there are $620$ possible choices.

Of course, you could also use $36(10)-100$ to avoid repetition, which yields the same answer.

Answer 2

Say thee are $d$ digits and $l$ letters. Yes, we all know $d=10$ and $l=26$, but forget about that for the moment. Now what is the problem's solution? It's $(d+l)^2-l^2=2dl+d^2$, as opposed to the $(d+l)d=dl+d^2$ you suggest. The extra factor of $2$ comes from not knowing which slot to choose for the digit, in the case there's one of them. There are also $d^2$ digit-only options. You're welcome to try a similar treatment of $n$-character results (so we've done $n=2$ before: the terms in $(d+l)^n-l^n$ are proportional to suitable binomial coefficients, reflecting the choice of slots for the digits.

Answer 3

36x10 would mean that the second character was always a number. This misses out 10x26 cases.

The solution is to take the number of cases with two alphanumeric characters (36x36) and subtract all those cases that don’t have at least one number - ie both letters (26x26). So $36*36-26*26 = 620$ and you get the same answer with $36*10 +10*26 =620$