Animated Solution for Mathematics - Permutations and Combinations: The value of $(2{\cdot }^1{P}_0-3{\cdot }^2{P}_1+4{\cdot }^3{P}_2-...\text{ up to }51^{th}\text{ term})+(1!-2!+3!-...\text{ up to }51^{th}\text{ term})$ is equal to :

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey to uncover the hidden elegance within a seemingly chaotic expression. When you first look at a problem like this in a JEE Advanced paper, your heart might skip a beat.

You see permutations, factorials, alternating signs, and a series that stretches out to fifty-one terms. It is designed to look intimidating. But remember, in the world of competitive mathematics, complexity is often just a mask for a beautiful, underlying simplicity.

We are presented with two distinct series, $S_1$ and $S_2$. Our objective is to find the sum $S=S_1+S_2$, defined as:

Do not try to calculate these terms individually. That is the trap. Instead, let us focus on the general term of the first series, $T_r$. By observing the pattern, we see that for the $r^{th}$ term, the multiplier is $(r+1)$ and the permutation is ${}^r{P}_{r-1}$.

Thus, the general term is $T_r=(r+1){\cdot }^r{P}_{r-1}$.

Now, let us apply our toolkit. Recall the definition of a permutation: ${}^n{P}_r=\frac{n!}{(n-r)!}$. If we substitute $n=r$ and the lower index as $r-1$, we get:

This is the 'Aha!' moment. The permutation, which looked complex, collapses into a simple factorial. Now, substitute this back into our general term $T_r$:

Using the fundamental property of factorials, where $(n+1)\cdot n!=(n+1)!$, we find that $T_r=(r+1)!$. Suddenly, the first series is not a collection of permutations anymore; it is a sequence of factorials: $2!-3!+4!-5!+\cdots +52!$.

Now, let us bring the two series together. This is where the magic happens. We have:

When we add $S_1$ and $S_2$, look at what happens to the terms. We have a positive $2!$ in $S_1$ and a negative $2!$ in $S_2$. They cancel out!

We have a negative $3!$ in $S_1$ and a positive $3!$ in $S_2$. They cancel out! This pattern continues, creating a 'telescoping' effect where almost every term is annihilated by its counterpart.

Everything in the middle vanishes into the void of zero. We are left with only the very first term of $S_2$ and the very last term of $S_1$.

After the dust settles, we are left with $S=1!+52!$. Since $1!=1$, our final answer is:

Look at how far we have come. We started with a terrifying expression involving permutations and alternating series, and through logical deconstruction and the application of fundamental properties, we reduced it to a simple addition. This is the essence of JEE Advanced mathematics.