Key Concept: Vandermonde's Identity

This problem primarily relies on Vandermonde's Identity, which is a fundamental identity in combinatorics used to evaluate sums of products of binomial coefficients. The identity states:

$\sum_{k=0}^{r} {^n C_k \cdot ^m C_{r-k}} = {^{n+m} C_r}$

This identity essentially tells us that if we choose $r$ items from a set of $n+m$ items, it's equivalent to choosing $k$ items from the first $n$ items and $r-k$ items from the remaining $m$ items, and summing over all possible values of $k$.

Another useful property of binomial coefficients is symmetry: ${^n C_r = ^n C_{n-r}}$.

Problem Setup

We are given the equation: $\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right) - \sum\limits_{k = 1}^{30} {\left( {{}^{30}{C_k}} \right)\left( {{}^{30}{C_{k - 1}}} \right) = {{\alpha (60!)} \over {(30!)(31!)}}} }$ Our goal is to find the value of $16\alpha$. We will evaluate each summation term separately.

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Step 1: Evaluating the First Summation

Let's evaluate the first sum: $S_1 = \sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right)}$.

• Expand the sum: $S_1 = {}^{31}{C_1} \cdot {}^{31}{C_0} + {}^{31}{C_2} \cdot {}^{31}{C_1} + {}^{31}{C_3} \cdot {}^{31}{C_2} + \dots + {}^{31}{C_{31}} \cdot {}^{31}{C_{30}}$

• Apply the symmetry property (${^n C_r = ^n C_{n-r}}$) to the second term in each product: We want to transform this sum into the form required by Vandermonde's Identity. To do this, we rewrite ${}^{31}{C_{k-1}}$ as ${}^{31}{C_{31-(k-1)}} = {}^{31}{C_{32-k}}$. However, a simpler approach for this specific sum is to rewrite the first term as ${}^{31}{C_k} = {}^{31}{C_{31-k}}$. This is often more straightforward when one index is fixed relative to the other. Let's change the second term: ${}^{31}{C_{k-1}} = {}^{31}{C_{31-(k-1)}} = {}^{31}{C_{32-k}}$. So the sum becomes: $S_1 = \sum\limits_{k = 1}^{31} {{}^{31}{C_k} \cdot {}^{31}{C_{32-k}}}$ This is not quite in the form of Vandermonde's Identity. Let's try rewriting the $C_{k-1}$ term from the original sum: Consider the term ${}^{31}{C_{k-1}}$. We can write this as ${}^{31}{C_{31-(k-1)}} = {}^{31}{C_{32-k}}$. So the sum is $\sum_{k=1}^{31} {^{31}C_k \cdot ^{31}C_{32-k}}$. Let $j = k-1$. Then $k = j+1$. When $k=1, j=0$. When $k=31, j=30$. The sum becomes: $S_1 = \sum_{j=0}^{30} {^{31}C_{j+1} \cdot ^{31}C_j}$ Now, apply symmetry to the second term: ${}^{31}{C_j} = {}^{31}{C_{31-j}}$. $S_1 = \sum_{j=0}^{30} {^{31}C_{j+1} \cdot {}^{31}{C_{31-j}}}$ This still doesn't look like the standard Vandermonde form directly.

  Let's follow the common technique from the given solution, which is simpler for this specific structure: Rewrite ${}^{31}{C_{k-1}}$ as ${}^{31}{C_{31-(k-1)}} = {}^{31}{C_{32-k}}$. Then: $S_1 = {}^{31}{C_1} \cdot {}^{31}{C_{31}} + {}^{31}{C_2} \cdot {}^{31}{C_{30}} + \dots + {}^{31}{C_{31}} \cdot {}^{31}{C_{1}}$ This is still not quite right.

  The key is to rewrite one of the terms such that the sum of the lower indices is constant. The existing solution used: ${}^{31}{C_k} \cdot {}^{31}{C_{k-1}}$ and transformed it into a sum like: ${}^{31}{C_0} \cdot {}^{31}{C_{30}} + {}^{31}{C_1} \cdot {}^{31}{C_{29}} + \dots + {}^{31}{C_{30}} \cdot {}^{31}{C_0}$ Let's derive this. We have ${}^{31}{C_k}$ and ${}^{31}{C_{k-1}}$. Apply ${}^n{C_r} = {}^n{C_{n-r}}$ to the second term: ${}^{31}{C_{k-1}} = {}^{31}{C_{31-(k-1)}} = {}^{31}{C_{32-k}}$. So the general term is ${}^{31}{C_k} \cdot {}^{31}{C_{32-k}}$. The sum is $\sum_{k=1}^{31} {^{31}C_k \cdot ^{31}C_{32-k}}$. Let $r = k$. Then the second index is $32-r$. The sum of the lower indices is $r + (32-r) = 32$. This implies that the sum is equal to ${}^{31+31}{C_{32}} = {}^{62}{C_{32}}$. Using the property ${^n C_r = ^n C_{n-r}}$, we have ${}^{62}{C_{32}} = {}^{62}{C_{62-32}} = {}^{62}{C_{30}}$.

• Applying Vandermonde's Identity (detailed): We have the sum $S_1 = \sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right)}$. To apply Vandermonde's Identity, we need one term with index $k$ and another with index $r-k$. Let's rewrite ${}^{31}{C_{k-1}}$ using symmetry: ${}^{31}{C_{k-1}} = {}^{31}{C_{31-(k-1)}} = {}^{31}{C_{32-k}}$. So, $S_1 = \sum\limits_{k = 1}^{31} {{}^{31}{C_k} \cdot {}^{31}{C_{32-k}}}$. Let $n=31$, $m=31$, and $r=32$. The sum is of the form $\sum_{k} {^n C_k \cdot ^m C_{r-k}}$. The minimum value of $k$ is $1$. The maximum value of $k$ is $31$. The minimum value of $32-k$ is $32-31=1$. The maximum value of $32-k$ is $32-1=31$. Since the binomial coefficients ${^N C_X}$ are zero if $X < 0$ or $X > N$, we can extend the summation range to $k=0$ to $k=32$ without changing the sum value. For $k=0$, ${}^{31}{C_0} \cdot {}^{31}{C_{32}}$ (since ${}^{31}{C_{32}} = 0$, this term is 0). For $k=32$, ${}^{31}{C_{32}} \cdot {}^{31}{C_0}$ (since ${}^{31}{C_{32}} = 0$, this term is 0). Therefore, $S_1 = \sum_{k=0}^{32} {^{31}C_k \cdot ^{31}C_{32-k}}$ Now, this exactly matches Vandermonde's Identity with $n=31$, $m=31$, and $r=32$. $S_1 = {}^{31+31}{C_{32}} = {}^{62}{C_{32}}$ Using the symmetry property ${^N C_X = ^N C_{N-X}}$: $S_1 = {}^{62}{C_{62-32}} = {}^{62}{C_{30}}$

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Step 2: Evaluating the Second Summation

Next, let's evaluate the second sum: $S_2 = \sum\limits_{k = 1}^{30} {\left( {{}^{30}{C_k}} \right)\left( {{}^{30}{C_{k - 1}}} \right)}$.

• Expand the sum: $S_2 = {}^{30}{C_1} \cdot {}^{30}{C_0} + {}^{30}{C_2} \cdot {}^{30}{C_1} + \dots + {}^{30}{C_{30}} \cdot {}^{30}{C_{29}}$

• Apply the symmetry property (${^n C_r = ^n C_{n-r}}$) to the second term in each product: Rewrite ${}^{30}{C_{k-1}}$ as ${}^{30}{C_{30-(k-1)}} = {}^{30}{C_{31-k}}$. So, $S_2 = \sum\limits_{k = 1}^{30} {{}^{30}{C_k} \cdot {}^{30}{C_{31-k}}}$.

• Applying Vandermonde's Identity (detailed): Let $n=30$, $m=30$, and $r=31$. The sum is of the form $\sum_{k} {^n C_k \cdot ^m C_{r-k}}$. The minimum value of $k$ is $1$. The maximum value of $k$ is $30$. The minimum value of $31-k$ is $31-30=1$. The maximum value of $31-k$ is $31-1=30$. Similar to $S_1$, we can extend the summation range to $k=0$ to $k=31$. For $k=0$, ${}^{30}{C_0} \cdot {}^{30}{C_{31}}$ (which is 0). For $k=31$, ${}^{30}{C_{31}} \cdot {}^{30}{C_0}$ (which is 0). Therefore, $S_2 = \sum_{k=0}^{31} {^{30}C_k \cdot ^{30}C_{31-k}}$ Now, this exactly matches Vandermonde's Identity with $n=30$, $m=30$, and $r=31$. $S_2 = {}^{30+30}{C_{31}} = {}^{60}{C_{31}}$ Using the symmetry property ${^N C_X = ^N C_{N-X}}$: $S_2 = {}^{60}{C_{60-31}} = {}^{60}{C_{29}}$

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Step 3: Substituting back into the Original Equation

Now we substitute the values of $S_1$ and $S_2$ back into the given equation: ${}^{62}{C_{30}} - {}^{60}{C_{29}} = {{\alpha (60!)} \over {(30!)(31!)}}$

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Step 4: Simplifying the Expression and Solving for $\alpha$

We need to express the binomial coefficients in terms of factorials and simplify to find $\alpha$. The right-hand side has a specific factorial form, so we aim to transform the left-hand side into that form as well.

• Express ${}^{62}{C_{30}}$ in terms of factorials: ${}^{62}{C_{30}} = \frac{62!}{30!(62-30)!} = \frac{62!}{30!32!}$ To match the denominator $(30!)(31!)$ on the right side, we can expand $62!$ and $32!$: ${}^{62}{C_{30}} = \frac{62 \times 61 \times 60!}{30! \times 32 \times 31!} = \left( \frac{62 \times 61}{32} \right) \frac{60!}{30!31!}$ This is a crucial step to make the terms comparable.

• Express ${}^{60}{C_{29}}$ in terms of factorials: ${}^{60}{C_{29}} = \frac{60!}{29!(60-29)!} = \frac{60!}{29!31!}$ To match the denominator $(30!)(31!)$ on the right side, we can multiply the numerator and denominator by $30$: ${}^{60}{C_{29}} = \frac{60!}{29!31!} = \frac{30 \times 60!}{30 \times 29!31!} = \frac{30 \times 60!}{30!31!}$

• Substitute these back into the equation: $\left( \frac{62 \times 61}{32} \right) \frac{60!}{30!31!} - \left( 30 \right) \frac{60!}{30!31!} = {{\alpha (60!)} \over {(30!)(31!)}}$

• Factor out the common term $\frac{60!}{30!31!}$: $\frac{60!}{30!31!} \left( \frac{62 \times 61}{32} - 30 \right) = {{\alpha (60!)} \over {(30!)(31!)}}$

• Cancel the common factorial term from both sides: $\frac{62 \times 61}{32} - 30 = \alpha$

• Calculate the value of $\alpha$: $\alpha = \frac{3782}{32} - 30$ $\alpha = \frac{1891}{16} - 30$ To combine, find a common denominator: $\alpha = \frac{1891}{16} - \frac{30 \times 16}{16} = \frac{1891 - 480}{16}$ $\alpha = \frac{1411}{16}$

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Step 5: Calculating $16\alpha$

The problem asks for the value of $16\alpha$. $16\alpha = 16 \times \frac{1411}{16}$ $16\alpha = 1411$

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Tips and Common Mistakes to Avoid

1. Careful Application of Vandermonde's Identity: Ensure the indices match the form $\sum_{k=0}^{r} {^n C_k \cdot ^m C_{r-k}}$. If they don't, use the symmetry property (${^n C_r = ^n C_{n-r}}$) to transform one of the terms. Also, be mindful of the summation limits; extending them to 0 or up to $n$ (or $m$) is often valid if the additional terms are zero.
2. Factorial Manipulation: When simplifying expressions involving factorials, aim to get a common factorial term that can be cancelled. This often involves expanding larger factorials (e.g., $N! = N \times (N-1)!$) or multiplying by terms like $\frac{X}{X}$ to adjust denominators.
3. Arithmetic Errors: Double-check calculations, especially when dealing with large numbers, multiplications, and subtractions.

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Summary

The problem was efficiently solved by recognizing and applying Vandermonde's Identity to simplify the two given summations. Each sum, after a simple transformation using the symmetry property of binomial coefficients, reduced to a single binomial coefficient. Subsequently, careful algebraic manipulation of the factorial forms allowed us to solve for $\alpha$ and ultimately find the value of $16\alpha$. The core takeaway is the power of Vandermonde's Identity in combinatorial sums.

The final answer is $\boxed{\text{1411}}$.