Calculating Sum of Binomial Coefficients

The problem asks us to find the value of the sum: \( C(47, 4) + C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3) + C(47, 3) \). This involves understanding and applying properties of binomial coefficients.

We can rearrange the terms in the sum to group similar terms and facilitate calculation:

Sum = \( C(47, 4) + C(47, 3) + C(48, 3) + C(49, 3) + C(50, 3) + C(51, 3) \)

A key property of binomial coefficients is Pascal's Identity, which states:

\( C(n, k) + C(n, k+1) = C(n+1, k+1) \)

Let's apply this identity step-by-step to the rearranged sum.

Step-by-Step Calculation using Pascal's Identity

We can combine the first two terms using Pascal's Identity with \( n = 47 \) and \( k = 3 \):

• \( C(47, 3) + C(47, 4) = C(47+1, 3+1) = C(48, 4) \)

Now the sum becomes:

Sum = \( C(48, 4) + C(48, 3) + C(49, 3) + C(50, 3) + C(51, 3) \)

Next, combine \( C(48, 4) \) and \( C(48, 3) \) using Pascal's Identity with \( n = 48 \) and \( k = 3 \):

• \( C(48, 3) + C(48, 4) = C(48+1, 3+1) = C(49, 4) \)

The sum is now:

Sum = \( C(49, 4) + C(49, 3) + C(50, 3) + C(51, 3) \)

Continue this process. Combine \( C(49, 4) \) and \( C(49, 3) \) with \( n = 49 \) and \( k = 3 \):

• \( C(49, 3) + C(49, 4) = C(49+1, 3+1) = C(50, 4) \)

The sum becomes:

Sum = \( C(50, 4) + C(50, 3) + C(51, 3) \)

Combine \( C(50, 4) \) and \( C(50, 3) \) with \( n = 50 \) and \( k = 3 \):

• \( C(50, 3) + C(50, 4) = C(50+1, 3+1) = C(51, 4) \)

The sum is now:

Sum = \( C(51, 4) + C(51, 3) \)

Finally, combine \( C(51, 4) \) and \( C(51, 3) \) with \( n = 51 \) and \( k = 3 \):

• \( C(51, 3) + C(51, 4) = C(51+1, 3+1) = C(52, 4) \)

Thus, the sum \( C(47, 4) + C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3) + C(47, 3) \) simplifies to \( C(52, 4) \).

Understanding Binomial Coefficients

A binomial coefficient, denoted as \( C(n, k) \) or \( \binom{n}{k} \), represents the number of ways to choose a subset of \( k \) elements from a set of \( n \) elements without regard to the order of selection. It is calculated using the formula:

\( C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

where \( n! \) is the factorial of \( n \), which is the product of all positive integers up to \( n \).

Key properties include:

• \( C(n, k) = C(n, n-k) \) (Symmetry Property)
• \( C(n, 0) = 1 \)
• \( C(n, n) = 1 \)
• \( C(n, 1) = n \)
• \( C(n, k) + C(n, k+1) = C(n+1, k+1) \) (Pascal's Identity)

Pascal's Identity is the property that was crucial in solving this problem by allowing us to combine adjacent terms in Pascal's triangle fashion.

Revision Table: Key Concepts

Additional Information: Hockey-stick Identity

The pattern observed in the step-by-step solution is related to a property called the Hockey-stick Identity (or Christmas Stocking Identity). This identity states that the sum of the binomial coefficients along a diagonal in Pascal's triangle is equal to the binomial coefficient below the last term on that diagonal.

The identity is formally written as:

\( \sum_{i=r}^{n} C(i, r) = C(n+1, r+1) \)

In our problem, we had the sum \( C(47, 4) + C(47, 3) + C(48, 3) + C(49, 3) + C(50, 3) + C(51, 3) \). We started by using Pascal's identity on \( C(47, 4) + C(47, 3) \). If we rewrite \( C(n,k) \) as \( C(n, n-k) \) using symmetry, the terms \( C(47, 3) + C(48, 3) + C(49, 3) + C(50, 3) + C(51, 3) \) are not directly in the form for the standard Hockey-stick identity summing over the upper index 'n' for a fixed lower index 'r'.

However, the step-by-step application of Pascal's Identity \( C(n, k) + C(n, k+1) = C(n+1, k+1) \) effectively creates a similar cumulative sum pattern. By starting with \( C(47, 4) + C(47, 3) \), we obtained \( C(48, 4) \), then adding \( C(48, 3) \) gives \( C(49, 4) \), and so on, until we reach \( C(52, 4) \).

This specific problem's structure allows for this iterative application of Pascal's Identity, leading to the final result.