Animated Solution for Mathematics - Probability: Let $S=\{1,2,3,4,5,6\}$. Then the probability that a randomly chosen onto function $g$ from $S$ to $S$ satisfies $g(3)=2g(1)$ is :

Imagine you are a detective tasked with solving a mystery involving a set of six numbers: $S=\{1,2,3,4,5,6\}$. We are looking for a specific type of function, $g:S\to S$, that is 'onto'.

In the world of JEE mathematics, when you see an 'onto' function between two finite sets of the same size, your intuition should immediately scream: 'Permutation!' Because the domain and codomain are identical, every element must map to a unique partner.

There is no room for overlap. This means our total sample space is simply the number of ways to arrange these six numbers, which is:

This is our universe of possibilities.

A Mathematical Anchor

The problem introduces a fascinating constraint: $g(3)=2g(1)$. This is not just an equation; it is a filter that limits our choices.

Since our codomain is limited to the set $S$, any output $g(x)$ must be between $1$ and $6$. This gives us a boundary condition for $g(1)$. If $g(1)$ were $4$, then $g(3)$ would be $8$, which is outside our set.

Thus, we must have $2g(1)\le 6$, which implies $g(1)\le 3$. This leaves us with only three possible scenarios for $g(1)$: $1,2,$ or $3$.

Let us break this down into three mutually exclusive cases.

If $g(1)=1$, then our rule dictates $g(3)=2(1)=2$. We have successfully locked in two mappings: $1\to 1$ and $3\to 2$.

Now, we have $4$ elements remaining in the domain and $4$ available spots in the codomain. Since the function must be a bijection, these remaining $4$ elements can be arranged in $4!$ ways:

If $g(1)=2$, then $g(3)=2(2)=4$. Again, we have fixed two mappings: $1\to 2$ and $3\to 4$.

The remaining $4$ elements are free to map to the remaining $4$ spots in the codomain. Just like before, this gives us:

Finally, if $g(1)=3$, then $g(3)=2(3)=6$. We have hit the boundary of our set.

We have fixed $1\to 3$ and $3\to 6$. The remaining $4$ elements can be arranged in $4!$ ways:

To find the total number of favorable outcomes, we simply sum the possibilities from our three cases:

Now, the probability is just the ratio of favorable outcomes to the total sample space:

When we simplify this fraction, we get the final result:

It is elegant, isn't it? By breaking a seemingly complex function problem into manageable, logical cases, we navigated the constraints and arrived at the solution. Remember, in JEE Advanced, the most complex problems are often just simple concepts disguised in layers of constraints.