Animated Solution for Mathematics - Permutations and Combinations: Let $A=[a_{ij}],a_{ij}\in \mathbb{Z}\cap [0,4],1\le i,j\le 2$. The number of matrices $A$ such that the sum of all entries is a prime number $p\in (2,13)$ is ______.

Imagine you are standing before a $2\times 2$ matrix, a simple grid of four cells. We label them $a,b,c,$ and $d$.

The problem constraints are deceptively simple: each entry must be an integer from the set $\{0,1,2,3,4\}$. Your task is to find how many such matrices exist where the sum of these four entries is a prime number $p$ in the range $(2,13)$.

At first glance, you might be tempted to start scribbling combinations. But stop. In the world of JEE Advanced, brute force is the enemy of elegance.

We are not just counting; we are solving a partition problem. We need a tool that handles the constraints of each variable being bounded between 0 and 4. That tool is the Generating Function.

For a single variable, the choices are represented by the polynomial $P(x)=1+x+x^2+x^3+x^4$. The exponent of $x$ represents the value chosen, and the coefficient represents the number of ways to choose it.

Since we have four independent variables ($a,b,c,d$), the total number of ways to achieve a sum $S$ is the coefficient of $x^S$ in the expansion of the product of these polynomials:

This is the heart of the problem. We are looking for the sum of the coefficients of $x^3,x^5,x^7,$ and $x^{11}$.

We can simplify our polynomial using the sum of a geometric progression. Notice that $1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$. Raising this to the power of 4, we get:

Now, let's expand the numerator using the binomial theorem:

We only care about terms up to $x^{11}$, so we can safely ignore $x^{15}$ and higher powers. For the denominator, we use the negative binomial expansion, where the coefficient of $x^r$ in $(1-x{)}^{-4}$ is $\left(\genfrac{}{}{0ex}{}{r+3}{3}\right)$.

Now, we extract the coefficients for each prime sum $p\in \{3,5,7,11\}$:

1. For $p=3$: We need the coefficient of $x^3$. This comes from $1\times (\text{coeff of }{x}^3\text{ in }(1-x{)}^{-4})$.

2. For $p=5$: We need the coefficient of $x^5$. This comes from $1\times (\text{coeff of }{x}^5)-4\times (\text{coeff of }{x}^0)$.

3. For $p=7$: We need the coefficient of $x^7$. This comes from $1\times (\text{coeff of }{x}^7)-4\times (\text{coeff of }{x}^2)$.

4. For $p=11$: We need the coefficient of $x^{11}$. This comes from $1\times (\text{coeff of }{x}^{11})-4\times (\text{coeff of }{x}^6)+6\times (\text{coeff of }{x}^1)$.

Adding these together, the total number of matrices is $20+52+80+52=204$.

While some answer keys might suggest 196, the rigorous mathematical derivation leads us to 204. Trust the process, trust the math, and you will always find the truth. You have just mastered a classic JEE Advanced combinatorial technique!