Key Concept: Vandermonde's Identity and Properties of Binomial Coefficients

The core of this problem lies in simplifying complex sums of products of binomial coefficients. The key mathematical tool for this is Vandermonde's Identity: $\sum_{k=0}^{r} \binom{m}{k} \binom{n}{r-k} = \binom{m+n}{r}$ This identity states that the number of ways to choose $r$ items from a set of $m+n$ items is equal to the sum of choosing $k$ items from $m$ and $r-k$ items from $n$, for all possible values of $k$.

Another crucial property of binomial coefficients used here is the symmetry property: $\binom{n}{k} = \binom{n}{n-k}$ This property allows us to transform binomial coefficients, which is often necessary to align terms with Vandermonde's Identity.

Step-by-Step Solution

1. Understanding the Given Notation The problem defines two notations:

• $\left( {\matrix{ n \cr k \cr } } \right)$ denotes the standard binomial coefficient ${}^n{C_k}$ or $\binom{n}{k}$.
• \left[ {\matrix{ n \cr k \cr } } \right] = \left\{ {\matrix{ {\left( {\matrix{ n \cr k \cr } } \right),} & {if\,0 \le k \le n} \cr {0,} & {otherwise} \cr } } \right.$$ This second notation simply formalizes that the binomial coefficient \binom{n}{k}$is$0$if$k$is out of the valid range (i.e.,$k < 0$or$k > n$). This is the standard convention for binomial coefficients and is important for extending the summation limits without changing the value of the sum.

The expression for $A_k$ is given as: ${A_k} = \sum\limits_{i = 0}^9 {\left( {\matrix{ 9 \cr i \cr } } \right)\left[ {\matrix{ {12} \cr {12 - k + i} \cr } } \right] + } \sum\limits_{i = 0}^8 {\left( {\matrix{ 8 \cr i \cr } } \right)\left[ {\matrix{ {13} \cr {13 - k + i} \cr } } \right]}$ Let's simplify each sum separately.

2. Simplifying the First Sum The first sum is $S_1 = \sum\limits_{i = 0}^9 {\left( {\matrix{ 9 \cr i \cr } } \right)\left[ {\matrix{ {12} \cr {12 - k + i} \cr } } \right]}$. To apply Vandermonde's Identity, we need the second binomial coefficient in the form $\binom{n}{r-i}$. We use the symmetry property $\binom{n}{k} = \binom{n}{n-k}$: $\left[ {\matrix{ {12} \cr {12 - k + i} \cr } } \right] = \binom{12}{12 - (12 - k + i)} = \binom{12}{k - i}$ Substituting this back into the sum: $S_1 = \sum\limits_{i = 0}^9 {\binom{9}{i}} \binom{12}{k - i}$ Now, this sum perfectly matches Vandermonde's Identity with $m=9$, $n=12$, and $r=k$. The summation limits naturally handle cases where $\binom{9}{i}$ or $\binom{12}{k-i}$ would be zero, effectively summing over all relevant $i$. Applying Vandermonde's Identity: $S_1 = \binom{9+12}{k} = \binom{21}{k}$

3. Simplifying the Second Sum The second sum is $S_2 = \sum\limits_{i = 0}^8 {\left( {\matrix{ 8 \cr i \cr } } \right)\left[ {\matrix{ {13} \cr {13 - k + i} \cr } } \right]}$. Again, we use the symmetry property to transform the second binomial coefficient: $\left[ {\matrix{ {13} \cr {13 - k + i} \cr } } \right] = \binom{13}{13 - (13 - k + i)} = \binom{13}{k - i}$ Substituting this back into the sum: $S_2 = \sum\limits_{i = 0}^8 {\binom{8}{i}} \binom{13}{k - i}$ This also matches Vandermonde's Identity, but this time with $m=8$, $n=13$, and $r=k$. Applying Vandermonde's Identity: $S_2 = \binom{8+13}{k} = \binom{21}{k}$

4. Combining the Terms to Find $A_k$ Now, we combine the simplified forms of $S_1$ and $S_2$ to find the expression for $A_k$: $A_k = S_1 + S_2 = \binom{21}{k} + \binom{21}{k}$ $A_k = 2 \binom{21}{k}$

5. Calculating $A_4 - A_3$ We need to find the value of $A_4 - A_3$. Substitute $k=4$ and $k=3$ into the expression for $A_k$: $A_4 = 2 \binom{21}{4}$ $A_3 = 2 \binom{21}{3}$ Therefore, $A_4 - A_3 = 2 \binom{21}{4} - 2 \binom{21}{3} = 2 \left( \binom{21}{4} - \binom{21}{3} \right)$

Next, we calculate the values of the binomial coefficients:

• For $\binom{21}{4}$: $\binom{21}{4} = \frac{21 \times 20 \times 19 \times 18}{4 \times 3 \times 2 \times 1}$ $= \frac{21 \times 20 \times 19 \times 18}{24}$ $= 21 \times 5 \times 19 \times 3 \quad (\text{since } 20/4 = 5 \text{ and } 18/(3 \times 2 \times 1) = 18/6 = 3)$ $= 105 \times 57 = 5985$
• For $\binom{21}{3}$: $\binom{21}{3} = \frac{21 \times 20 \times 19}{3 \times 2 \times 1}$ $= \frac{21 \times 20 \times 19}{6}$ $= 7 \times 10 \times 19 \quad (\text{since } 21/3 = 7 \text{ and } 20/2 = 10)$ $= 70 \times 19 = 1330$

Now, substitute these values back: $A_4 - A_3 = 2 (5985 - 1330)$ $A_4 - A_3 = 2 (4655)$ $A_4 - A_3 = 9310$

6. Solving for $p$ The problem states that $A_4 - A_3 = 190p$. We have calculated $A_4 - A_3 = 9310$. So, we set up the equation: $190p = 9310$ Now, solve for $p$: $p = \frac{9310}{190}$ $p = \frac{931}{19}$ To perform the division: $931 \div 19$: $19 \times 4 = 76$. $93 - 76 = 17$. Bring down the $1$, making it $171$. $19 \times 9 = 171$. So, $p = 49$.

Tips and Common Mistakes to Avoid

• Misinterpreting Notation: Always clarify any non-standard notation at the beginning. Here, the bracket notation for binomial coefficients simply means the standard definition.
• Incorrect Application of Vandermonde's Identity: Ensure that the indices correctly align with the identity. Specifically, check that the sum of the lower indices of the binomial coefficients in each term is constant (e.g., $i + (k-i) = k$).
• Forgetting Symmetry Property: The property $\binom{n}{k} = \binom{n}{n-k}$ is frequently used to manipulate terms into a form suitable for Vandermonde's Identity or other binomial identities.
• Calculation Errors: Binomial coefficient calculations, especially with larger numbers, are prone to arithmetic errors. Double-check your multiplication and division.
• Summation Limits: Remember that $\binom{n}{k}=0$ if $k < 0$ or $k > n$. This implies that the explicit summation limits (e.g., $i=0$ to $9$) are generally sufficient, as any terms outside the practical range of $i$ for which $\binom{12}{k-i}$ is non-zero will automatically be zero due to the $\binom{9}{i}$ term, or vice-versa.

Summary and Key Takeaway This problem is an excellent illustration of how to simplify complex summations involving binomial coefficients by recognizing and applying Vandermonde's Identity. The key steps involved:

1. Understanding and interpreting the given notation.
2. Using the symmetry property of binomial coefficients ($\binom{n}{k} = \binom{n}{n-k}$) to transform terms into a suitable form.
3. Applying Vandermonde's Identity to reduce the summations to single binomial coefficients.
4. Carefully calculating the resulting binomial coefficients.
5. Solving the final algebraic equation. Mastering these binomial identities is crucial for solving many problems in combinatorics and probability, frequently encountered in competitive exams like JEE.