If 1P1 + 2 * 2P2 + 3 * 3P3 + ... + 15 * 15P15 = qPr - s, 0 ≤ s ≤ 1 then q+sCr-s is equal to .

Q: Wait, why is $^kP_k$ equal to $k!$?

Great question! By definition, $^nP_r = \frac{n!}{(n-r)!}$. When $n=r$, we get $^nP_n = \frac{n!}{(n-n)!} = \frac{n!}{0!}$. Since $0!$ is defined as $1$, $^nP_n$ simply becomes $n!$. It is a fundamental property that you should keep in your toolkit!

Q: How did you know to rewrite $k$ as $(k+1-1)$?

That is the 'telescopic trick'! When you see a term like $k \cdot k!$, you want to transform it into a form $f(k+1) - f(k)$ so that when you sum it, most terms cancel out. Adding and subtracting 1 is a standard algebraic maneuver to force that structure. Always look for ways to create a difference of consecutive terms in series problems.

Q: Why does the telescoping sum leave only the first and last terms?

Imagine a line of dominoes. When you expand $\sum_{k=1}^{15} ((k+1)! - k!)$, you get $(2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + (16! - 15!)$. Notice that the $+2!$ from the first term cancels the $-2!$ from the second, the $+3!$ cancels the $-3!$, and so on. Only the very first negative term ($-1!$) and the very last positive term ($16!$) survive the carnage!

Question

If $^1P_1 + 2 \cdot ^2P_2 + 3 \cdot ^3P_3 + \dots + 15 \cdot ^{15}P_{15} = ^qP_r - s, 0 \leq s \leq 1$ then $^{q+s}C_{r-s}$ is equal to .

Sigma Percentile · Accepted Answer

### The Beauty of Telescoping Series

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to unravel a problem that, at first glance, looks like a mountain of factorials and permutations.

But as we peel back the layers, you will see that it is actually a beautifully choreographed dance of numbers. Let us dive in.

### Phase 1: Decoding the General Term

We are presented with the series $S = ^1P_1 + 2 \cdot ^2P_2 + 3 \cdot ^3P_3 + \dots + 15 \cdot ^{15}P_{15}$. The first step in any series problem is to find the heartbeat of the sequence—the general term, $T_k$.

Looking at the pattern, it is clear that $T_k = k \cdot ^kP_k$. Now, recall the definition of a permutation: $^nP_r = \frac{n!}{(n-r)!}$.

When we apply this to $^kP_k$, we get:

$$T_k = k \cdot \frac{k!}{(k-k)!} = \frac{k \cdot k!}{0!}$$

Since $0! = 1$, our permutation term simplifies elegantly to $^kP_k = k!$. Suddenly, our general term becomes $T_k = k \cdot k!$.

This is much cleaner, but we still have a product of $k$ and $k!$, which is not immediately summable.

### Phase 2: The Telescopic Magic

This is where we pull a classic JEE trick from our arsenal. We need to transform $T_k = k \cdot k!$ into something that allows for a 'telescopic' cancellation.

The trick is to manipulate the $k$ factor by writing $k$ as $(k+1-1)$. This allows us to distribute the $k!$ term:

$$T_k = (k+1-1) \cdot k! = (k+1) \cdot k! - 1 \cdot k!$$

Now, look at the first part: $(k+1) \cdot k!$. By the definition of factorials, this is exactly $(k+1)!$.

So, our general term transforms into:

$$T_k = (k+1)! - k!$$

We have successfully turned a product into a difference of two consecutive factorials. This is the key to the entire problem.

### Phase 3: The Domino Effect

Now, let us expand the summation $S = \sum_{k=1}^{15} ((k+1)! - k!)$. When we write out the terms, the magic happens:

$$S = (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + (16! - 15!)$$

Look closely at the diagonal cancellation. The positive $2!$ from the first term cancels the negative $2!$ from the second. The $3!$ cancels out, the $4!$ cancels out, and this domino effect continues all the way to the end.

The only terms left standing are the very first negative term, $-1!$, and the very last positive term, $16!$. Thus, our massive sum collapses to:

$$S = 16! - 1$$

### Phase 4: The Final Comparison and Calculation

The problem states that $S = ^qP_r - s$, where $0 \leq s \leq 1$. We have found $S = 16! - 1$.

Since $16!$ is equivalent to $^{16}P_{16}$, we can write our sum as $S = ^{16}P_{16} - 1$. By comparing this to the given form, we identify $q = 16$, $r = 16$, and $s = 1$.

The final task is to calculate $^{q+s}C_{r-s}$, which is $^{16+1}C_{16-1} = ^{17}C_{15}$. Using the symmetry property of combinations:

$$^{17}C_{15} = ^{17}C_{17-15} = ^{17}C_2$$

This is a simple calculation:

$$\frac{17 	imes 16}{2 	imes 1} = 17 	imes 8 = 136$$

And there you have it—a complex series reduced to an elegant, simple integer. Remember, in JEE Advanced, the most intimidating problems often have the most beautiful, simple solutions hidden just beneath the surface.

**The final answer is 136.**

Animated Solution for Mathematics - Permutations and Combinations: If $^{1} P_{1} + 2 \cdot^{2} P_{2} + 3 \cdot^{3} P_{3} + \dots + 15 \cdot^{15} P_{15} =^{q} P_{r} - s, 0 \leq s \leq 1$ then $^{q + s} C_{r - s}$ is equal to .

Enter Numerical Value:

Visualized Solution

Identifying the General Term

Converting Permutation to Factorial

Updating the General Term

The Telescopic Trick

Distributing the Factorial

Expanding the Summation

Telescoping Cancellation

Comparing with the Given Equation

Final Substitution

Calculating the Final Answer

The Sigma Insight: Linear Permutations

The Beauty of Telescoping Series

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Permutations and Combinations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1P1​+2⋅2P2​+3⋅3P3​+⋯+15⋅15P15​=qPr​−s,0≤s≤1 then q+sCr−s​ is equal to .