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

The given summation is:

\( \sum_{k=0}^6 \binom{51-k}{3} \)

Step 1: Rewrite the summation

This summation can be expanded as:

\( \sum_{k=0}^6 \binom{51-k}{3} = \binom{51}{3} + \binom{50}{3} + \dots + \binom{45}{3}. \)

This is a finite summation of combinations.

Step 2: Use the telescoping property of combinations

We utilize the following identity for summation of combinations:

\( \sum_{r=a}^b \binom{r}{p} = \binom{b+1}{p+1} - \binom{a}{p+1} \)

Here, let \( a=45, b=51\), and \(p=3\). Substituting these values:

\( \sum_{k=0}^6 \binom{51-k}{3} = \binom{52}{4} - \binom{45}{4}. \)

Step 3: Verify the answer

From the above calculation, we see that:

\( \sum_{k=0}^6 \binom{51-k}{3} = \binom{52}{4} - \binom{45}{4}. \)

This matches option (3).

Final Answer:

\( \boxed{\binom{52}{4} - \binom{45}{4}} \)

Answer 2

The correct answer is (B) : ${ }^{52} C _4-{ }^{45} C _4$
$\sum _{k=0}^6{}^{51-k}{C}_3$
$={}^{51}{C}_3+{}^{50}{C}_3+{}^{49}{C}_3+\dots +{}^{45}{C}_3$
$={}^{45}{C}_3+{}^{46}{C}_3+\dots ..+{}^{51}{C}_3$
$={}^{45}{C}_4+{}^{45}{C}_3+{}^{46}{C}_3+\dots ..+{}^{51}{C}_3-{}^{45}{C}_4$
$\left({}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r\right)$
$={}^{52}{C}_4-{}^{45}{C}_4$