Let \(\binom{n}{k}\) denote \(^nC_k\).
If \(A_k = \sum_{i=0}^{9} \binom{9}{i} \binom{12}{12-k+i} + \sum_{i=0}^{8} \binom{8}{i} \binom{13}{13-k+i}\) and \(A_4 - A_3 = 190 p\), then p is equal to ___________.
Show Hint
The identity \(\binom{n}{r} = \binom{n}{n-r}\) is often the key to transforming a sum of products of binomial coefficients into a form where Vandermonde's Identity can be applied. Always check if this transformation simplifies the problem.
Step 1: Understanding the Question
We are given an expression \(A_k\) which is a sum of two series involving binomial coefficients. We need to evaluate \(A_4 - A_3\) and solve for \(p\). Step 2: Key Formula or Approach
We will use the identity \(\binom{n}{r} = \binom{n}{n-r}\) and Vandermonde's Identity, which states that the coefficient of \(x^k\) in \((1+x)^m(1+x)^n\) is \(\binom{m+n}{k}\), leading to the summation formula \(\sum_{i=0}^{k} \binom{m}{i}\binom{n}{k-i} = \binom{m+n}{k}\). Step 3: Detailed Explanation
Let's simplify the first sum in \(A_k\). Let it be \(S_1\).
\[ S_1 = \sum_{i=0}^{9} \binom{9}{i} \binom{12}{12-k+i} \]
Using \(\binom{n}{r} = \binom{n}{n-r}\), we have \(\binom{12}{12-k+i} = \binom{12}{12 - (12-k+i)} = \binom{12}{k-i}\).
\[ S_1 = \sum_{i=0}^{9} \binom{9}{i} \binom{12}{k-i} \]
This is the coefficient of \(x^k\) in the expansion of \((1+x)^9 (1+x)^{12} = (1+x)^{21}\).
By Vandermonde's Identity, \(S_1 = \binom{9+12}{k} = \binom{21}{k}\).
Now let's simplify the second sum, \(S_2\).
\[ S_2 = \sum_{i=0}^{8} \binom{8}{i} \binom{13}{13-k+i} \]
Using \(\binom{n}{r} = \binom{n}{n-r}\), we have \(\binom{13}{13-k+i} = \binom{13}{13 - (13-k+i)} = \binom{13}{k-i}\).
\[ S_2 = \sum_{i=0}^{8} \binom{8}{i} \binom{13}{k-i} \]
This is the coefficient of \(x^k\) in the expansion of \((1+x)^8 (1+x)^{13} = (1+x)^{21}\).
By Vandermonde's Identity, \(S_2 = \binom{8+13}{k} = \binom{21}{k}\).
So, \(A_k = S_1 + S_2 = \binom{21}{k} + \binom{21}{k} = 2\binom{21}{k}\).
Now we need to calculate \(A_4 - A_3\).
\[ A_4 = 2\binom{21}{4} \quad \text{and} \quad A_3 = 2\binom{21}{3} \]
\[ A_4 - A_3 = 2\left(\binom{21}{4} - \binom{21}{3}\right) \]
\(\binom{21}{3} = \frac{21 \times 20 \times 19}{3 \times 2 \times 1} = 7 \times 10 \times 19 = 1330\).
\(\binom{21}{4} = \frac{21 \times 20 \times 19 \times 18}{4 \times 3 \times 2 \times 1} = 21 \times 5 \times 19 \times 3 = 5985\).
\[ A_4 - A_3 = 2(5985 - 1330) = 2(4655) = 9310 \]
We are given that \(A_4 - A_3 = 190 p\).
\[ 9310 = 190 p \]
\[ p = \frac{9310}{190} = \frac{931}{19} = 49 \]
Step 4: Final Answer
The value of p is 49.
