Counting 4-digit combinations such that the first digit is positive and even, second is prime, third is Fibonacci, and fourth is triangular This seemed like a basic problem, but for some reason I can't figure it out:

In a $4$-digit combination, the first digit has to be a positive even number, the second a prime number, the third a Fibonacci number and the fourth a triangular number. The question asks me how many unique sets of $4$ combinations can I select from if digits CAN be repeated.
Answer choices:

*

*(A) $625$

*(B) $375$

*(C) $300$

*(D) $256$

*(E) $240$

This was my attempt:
All the numbers have to be less than $10$ as I need one digit. The sets which each digit can come from are

*

*Even number = $\{2,4,6,8\}$

*Prime number = $\{2,3,5,7\}$

*Fibonacci number (quite confused!!) = $\{0,1,1,2,3,5,8\}$ or just one $1$ ?

*Triangular number = $\{1,3,6\} $
Then perhaps the total number of such numbers is given by $4 \cdot 4 \cdot 7 \cdot 3 = 336.$ But that's not an answer choice! Or, with one less Fibonacci number, $4 \cdot 4 \cdot 6 \cdot 3 = 288$, which is still not a choice.
So I am totally confused and would appreciate any help.

P.S. The question is from MATH UIL (2014 Regionals) for anyone wondering. Go to page 49 in this PDF
(The question asks: 13. Willie Lawkette uses...)
 A: I'm going to make two comments on potential ambiguities in your question - the likely source of your questions and that give anyone a problem in solving this. It shows that the problem is very poorly framed if you're giving a multiple-choice exam.


Fibonacci number (Quite confused!!) = $\{0,1,1,2,3,5,8\}$ or just one $1$ ?

For this, note that only having one $1$ is relevant. Think about what the four-digit number would look like if you chose a $1$ from a pair of $1$'s - sure, different "numbers" in some sense, but the four-digit number would be the same regardless of which one you get.
In that sense, it is more fruitful to think about distinct or unique four digit combinations.
Also, a number is a Fibonacci number if it appears at all in the Fibonacci sequence, i.e. $1$ is not somehow "twice as much" a Fibonacci number (whatever that would mean) as the others. Basically the question you ask yourself is "is this number in the Fibonacci sequence?" If so, include it in the set. If not, don't.
Now, an issue: is $0$ a Fibonacci number?
Well, consider how we define the $n$-th Fibonacci number:
$$F_n = F_{n-1} + F_{n-2}$$
But for this to be useful, we need to define some "first" values, some seed values. We can define $F_0 = 0, F_1 = 1$. Or sometimes we define $F_1 = F_0 = 1$. Both sequences are ultimately the same, there's just a shifting over of the terms. Notice how one definition explicitly forbids $0$ from being in the sequence unless we work backwards (which is not the "standard" Fibonacci sequence in a sort of sense).


Triangular number = $\{1,3,6\}$

Herein lies the second ambiguity I wish to comment on.
A definition for the $n$-th triangular number $T_n$ can be given by
$$T_n = \sum_{k=0}^n k = 0 + 1 + 2 + ... + n = \frac{n(n+1)}{2}$$
Take $n=0$. The sum is zero, giving $0$ as the "zeroth" triangular number.
This definition is essentially the same as on Wikipedia (https://en.wikipedia.org/wiki/Triangular_number) except we start summing at $0$ and not $1$. This is perfectly fine though.

So, what have we noticed in this discussion? Three main things:


*

*First, you're not going to include $1$ twice in the set for that digit.

*Is $0$ a Fibonacci number? You can define it as the seed, or not, without meaningfully affecting the sequence.

*Is $0$ a triangular number? You can define $T_n$ in a manner which permits it to be one without issue.


I think we can claim that there are four choices for each of the first two digits in your four-digit number without issue. But we might have $5$ or $6$ for the third, and $3$ or $4$ for the fourth.
So in that light, we could have 
$$4 \cdot 4 \cdot 5 \cdot 3 = 240$$
$$4 \cdot 4 \cdot 5 \cdot 4 = 320$$
$$4 \cdot 4 \cdot 6 \cdot 3 = 288$$
$$4 \cdot 4 \cdot 6 \cdot 4 = 384$$
Now, luckily, only the first corresponds to an answer (E) - that being, $0$ is neither a triangular number nor a Fibonacci number.
But that these definitions so easily permit $0$ as being either and some places will cite it as being such. For example, per Wikipedia (https://en.wikipedia.org/wiki/Fibonacci_number), $0$ is a Fibonacci number is more recent books, but typically omitted in older texts. The OEIS sequence for triangular numbers (https://oeis.org/A000217) starts at the "zeroth" one, $0$.
So I feel with these ambiguities in place the question wasn't properly framed, but that's perhaps a matter of opinion.
I guess one could make the argument that to be triangular or Fibonacci, a number must be natural. (But then that touches on the debate of whether $0$ is a natural number, doesn't it? So being positive would likely be less ambiguous. :p).
So I suppose I'll end this post by giving you those things to mull over.
