Animated Solution for Mathematics - Functions: The number of one-one function $f:\{a,b,c,d\}\to \{0,1,2,\dots ,10\}$ such that $2f(a)-f(b)+3f(c)+f(d)=0$ is ______.

My dear student, welcome to a beautiful puzzle. In the world of JEE Advanced, we often encounter problems that seem like a chaotic mess of possibilities.

We have a function $f:\{a,b,c,d\}\to \{0,1,2,\dots ,10\}$. We are told it is a one-one function, meaning no two elements in the domain map to the same element in the codomain.

And then, we are hit with this: $2f(a)-f(b)+3f(c)+f(d)=0$. At first glance, this looks like a nightmare of combinations. But let's breathe and peel back the layers. The secret to solving this is not brute force; it is strategic containment.

The Rearrangement

The equation $2f(a)-f(b)+3f(c)+f(d)=0$ is just a relationship waiting to be tamed. Let's isolate the variable that is being subtracted:

Now, look at the right side. We know that $f(b)$ must be an element of our codomain, so $f(b)\le 10$.

This is our anchor! It means the entire sum $2f(a)+3f(c)+f(d)$ must be less than or equal to $10$. This simple realization turns a massive problem into a manageable one.

The Pivot Strategy

How do we count these without losing our minds? We look for the 'heavy hitter'—the term that restricts our options the fastest.

Here, that is $3f(c)$. Because its coefficient is $3$, it will hit that limit of $10$ much faster than the others.

So, we pivot our entire analysis around $f(c)$. We will test values for $f(c)$ starting from $0$ and see where the boundary stops us.

The Case Analysis

Let's walk through the cases together.

If $f(c)=0$, our equation simplifies to $2f(a)+f(d)=f(b)$. We need to pick distinct values for $f(a),f(d),$ and $f(b)$ from the set $\{1,2,\dots ,10\}$ (since $0$ is taken by $f(c)$).

By testing $f(a)=1,2,3,4$, we find $7+5+3+2=17$ valid functions. It is a beautiful, systematic dance of numbers.

If $f(c)=1$, the equation becomes $2f(a)+f(d)+3=f(b)$. Again, we test values for $f(a)$, ensuring we don't reuse $1$ or $f(c)$. This yields $9$ valid functions.

If $f(c)=2$, we get $2f(a)+f(d)+6=f(b)$. This is getting tighter! We find $4$ valid functions.

If $f(c)=3$, we get $2f(a)+f(d)+9=f(b)$. The room for error is almost gone. Only $1$ function survives this constraint.

The Final Summation

Why stop at $f(c)=3$? Because if $f(c)=4$, then $3f(c)=12$, which is already greater than our maximum allowed value of $10$. The math itself tells us when to stop.

Finally, we sum our cases: $17+9+4+1=31$.

There you have it. We didn't just calculate a number; we navigated a logical landscape. Remember, in JEE, the most complex problems are often just simple constraints waiting for the right perspective. The final answer is 31.