Question

\(\displaystyle\sum_{k=0}^6{ }^{51-k} C_3\) is equal to

• A. ${ }^{51} C _3-{ }^{45} C _3$
• B. ${ }^{52} C _4-{ }^{45} C _4$
• C. ${ }^{52} C _3-{ }^{45} C _3$
• D. ${ }^{51} C _4-{ }^{45} C _4$

Answer 1

The Correct Option is B

To solve the given problem, we need to evaluate the summation:

\(\displaystyle\sum_{k=0}^6{ }^{51-k} C_3\)

First, let's break down the expression inside the summation:

• When \(k = 0\), we have \({}{^{51} C_3}\)
• When \(k = 1\), we have \({}{^{50} C_3}\)
• Continue this pattern up to \(k = 6\):
• When \(k = 6\), we have \({}{^{45} C_3}\)

The expression becomes:

\({}{^{51} C_3 + ^{50} C_3 + ^{49} C_3 + ^{48} C_3 + ^{47} C_3 + ^{46} C_3 + ^{45} C_3}\)

Recognize that the sum of binomial coefficients of the type \({}{^{n} C_r}\) is a standard result in combinatorics, utilizing the Hockey Stick Identity (also known as the Christmas Stocking Theorem), which states:

\(\sum_{i=r}^{n} {^{i} C_r} = {^{n+1} C_{r+1}}\)

Applying this identity to our problem:

• Here \(n = 51\), \(k = 6\), \(r = 3\).
• According to the theorem, we have:
• \(\sum_{i=45}^{51} {^{i} C_3} = {^{52} C_4}\)

Now, only \({}{^{45} C_3}\) is left from the summation rule, because it was the lower bound of our sum:

Thus, the entire expression evaluates as:

\({}{^{52} C_4 - ^{45} C_3}\). Unfortunately, the form \({}{^{45} C_3}\) should not be subtracted independently here, meaning we have actually \({}{^{52} C_4 - ^{45} C_4}\) form difference, coming back to identity correctness.

This modifies and confirms the answer option:

The correct answer is: \({}{^{52} C_4 - ^{45} C_4}\)