Animated Solution for Mathematics - Binomial Theorem: The value of ${\sum }_{r=0}^6({}^6{C}_r\cdot {}^6{C}_{6-r})$ is equal to :

Unlocking the Binomial Sum

Have you ever looked at a problem and felt the urge to start calculating immediately? I know that feeling.

You see ${\sum }_{r=0}^6({}^6{C}_r\cdot {}^6{C}_{6-r})$ and your instinct is to start writing out ${}^6{C}_0\cdot {}^6{C}_6+{}^6{C}_1\cdot {}^6{C}_5+\dots$. While that is a valid path, it is a long, winding road filled with potential for arithmetic slips.

Today, we are going to take the shortcut—the path of the mathematician.

Let us pause and look at the structure. We have a product of two binomial coefficients.

Look at the lower indices: $r$ and $6-r$. When you add them together, what do you get? $r+(6-r)=6$.

This is not a coincidence. It is a beautiful, hidden symmetry. In the world of combinatorics, whenever you see a sum of products where the lower indices add up to a constant, you are likely looking at a story about selection.

Imagine you have two baskets. Basket A contains 6 red balls, and Basket B contains 6 blue balls. You are tasked with picking a total of 6 balls from these two baskets.

How many ways can you do this?

Well, you could pick 0 from Basket A and 6 from Basket B. That is ${}^6{C}_0\cdot {}^6{C}_6$. Or you could pick 1 from Basket A and 5 from Basket B. That is ${}^6{C}_1\cdot {}^6{C}_5$.

If you continue this logic, you realize that the sum ${\sum }_{r=0}^6({}^6{C}_r\cdot {}^6{C}_{6-r})$ is simply the total number of ways to choose 6 balls from the combined pool of 12 balls (6 red + 6 blue).

This realization is the heart of Vandermonde's Identity. It tells us that choosing $k$ items from two combined groups of size $n$ and $m$ is exactly the same as choosing $k$ items from a single group of size $n+m$.

Mathematically, we have collapsed our complex summation into a single, elegant term:

Isn't that breathtaking? We have turned a series of seven calculations into one.

Now, we just need to evaluate ${}^{12}{C}_6$. Using the standard formula ${}^n{C}_k=\frac{n!}{k!(n-k)!}$, we get:

Let's simplify this with care. We don't want to multiply huge numbers.

Notice that $6\times 2=12$, so we can cancel the $12$ in the numerator. Then, $5\times 4=20$, and $10\times 8=80$. Since $80/20=4$, we have simplified significantly.

Finally, $9/3=3$. We are left with $11\times 3\times 4\times 7$. Multiplying these gives us $33\times 28=924$.

Whenever you face a summation in JEE, don't rush to calculate. Ask yourself: 'Is there a combinatorial story here?'

Look for the symmetry, look for the identity, and let the math do the heavy lifting for you. You have just mastered a core concept that separates the calculators from the thinkers. Keep that curiosity alive!