Key Concept: Vandermonde's Identity

The problem asks us to evaluate a sum involving products of binomial coefficients. This specific pattern is directly related to Vandermonde's Identity.

Vandermonde's Identity states that for non-negative integers $m$, $n$, and $r$: $\sum_{k=0}^r \binom{m}{k} \binom{n}{r-k} = \binom{m+n}{r}$ This identity can be understood by considering the coefficient of $x^r$ in the expansion of $(1+x)^m (1+x)^n = (1+x)^{m+n}$. When we multiply the expansions of $(1+x)^m$ and $(1+x)^n$, the terms contributing to $x^r$ are formed by multiplying $x^k$ from the first expansion with $x^{r-k}$ from the second expansion. The coefficient of such a term would be $\binom{m}{k} \binom{n}{r-k}$. Summing these for all possible values of $k$ (from $0$ to $r$) gives the total coefficient of $x^r$ in $(1+x)^{m+n}$, which is $\binom{m+n}{r}$.

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Step-by-Step Solution

1. Analyze the Given Sum The given sum is: $S = \sum\limits_{r = 0}^6 {\left( {{}^6{C_r}\,.\,{}^6{C_{6 - r}}} \right)}$ Let's write out a few terms to observe the pattern:

• When $r=0$: ${}^6{C_0}\,.\,{}^6{C_{6 - 0}} = {}^6{C_0}\,.\,{}^6{C_6}$
• When $r=1$: ${}^6{C_1}\,.\,{}^6{C_{6 - 1}} = {}^6{C_1}\,.\,{}^6{C_5}$
• When $r=2$: ${}^6{C_2}\,.\,{}^6{C_{6 - 2}} = {}^6{C_2}\,.\,{}^6{C_4}$
• ...
• When $r=6$: ${}^6{C_6}\,.\,{}^6{C_{6 - 6}} = {}^6{C_6}\,.\,{}^6{C_0}$

So the sum is: $S = {}^6{C_0}{}^6{C_6} + {}^6{C_1}{}^6{C_5} + {}^6{C_2}{}^6{C_4} + {}^6{C_3}{}^6{C_3} + {}^6{C_4}{}^6{C_2} + {}^6{C_5}{}^6{C_1} + {}^6{C_6}{}^6{C_0}$ Explanation: We expand the summation to clearly see the individual terms involved. This helps in recognizing the structure that matches Vandermonde's Identity. Notice that for each term, the sum of the lower indices is $r + (6-r) = 6$.

2. Apply Vandermonde's Identity Comparing our sum $S$ with the general form of Vandermonde's Identity $\sum_{k=0}^r \binom{m}{k} \binom{n}{r-k} = \binom{m+n}{r}$:

• The first binomial coefficient is ${}^6{C_r}$, so $m=6$ and $k=r$.
• The second binomial coefficient is ${}^6{C_{6-r}}$, so $n=6$ and $r-k = 6-r$, which means the target power of $x$ (denoted as $r$ in the identity) is also $6$.
• The summation runs from $k=0$ to $k=6$, which matches the range of $r$.

Therefore, substituting $m=6$, $n=6$, and $r=6$ into Vandermonde's Identity: $S = \binom{6+6}{6} = \binom{12}{6}$ Explanation: We directly apply Vandermonde's Identity because the structure of the given sum perfectly matches the identity's form, simplifying a complex summation into a single binomial coefficient.

3. Calculate the Binomial Coefficient Now we need to calculate the value of ${}^{12}{C_6}$: ${}^{12}{C_6} = \frac{12!}{6!(12-6)!} = \frac{12!}{6!6!}$ ${}^{12}{C_6} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6!}{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 6!}$ We can cancel out $6!$ from the numerator and denominator: ${}^{12}{C_6} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$ Now, let's simplify the terms:

• $6 \times 2 = 12$, so $\frac{12}{6 \times 2} = 1$.
• $10/5 = 2$.
• $9/3 = 3$.
• $8/4 = 2$.

Substituting these simplifications: ${}^{12}{C_6} = 1 \times 11 \times 2 \times 3 \times 2 \times 7$ ${}^{12}{C_6} = 11 \times (2 \times 3 \times 2 \times 7) = 11 \times (6 \times 14) = 11 \times 84$ ${}^{12}{C_6} = 924$ Explanation: This step involves the direct calculation of the binomial coefficient using its definition. Carefully simplifying the factorial expression reduces the chance of arithmetic errors.

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Tips and Common Mistakes

• Recognizing Patterns: Always look for patterns in sums of binomial coefficients. If you see products of coefficients where the upper indices are constant ($n$ and $m$) and the lower indices sum to a constant ($k + (r-k) = r$), it's a strong indicator for Vandermonde's Identity or the coefficient extraction method.
• Coefficient Extraction Method: The problem can also be thought of as finding the coefficient of $x^6$ in the expansion of $(1+x)^6 \cdot (1+x)^6$.
  • We know $(1+x)^6 = {}^6C_0 + {}^6C_1 x + \dots + {}^6C_6 x^6$.
  • The product $(1+x)^6 \cdot (1+x)^6 = (1+x)^{12}$.
  • The coefficient of $x^6$ in $(1+x)^{12}$ is ${}^{12}C_6$. This method essentially derives Vandermonde's Identity in a specific case and leads to the same result.
• Common Mistake: Sometimes students confuse $\sum_{r=0}^n \left( {}^n{C_r} \cdot {}^n{C_{n-r}} \right)$ with $\sum_{r=0}^n \left( {}^n{C_r} \right)^2$. While both result in ${}^{2n}{C_n}$, it's important to understand why. The property ${}^n{C_k} = {}^n{C_{n-k}}$ makes them equivalent in value, but the identity ${}^n{C_r} \cdot {}^n{C_{n-r}}$ is the direct form for Vandermonde's identity with $m=n$ and $r=n$.

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Summary

The problem demonstrates a classic application of Vandermonde's Identity (or the coefficient extraction technique from polynomial multiplication). By recognizing the sum of products of binomial coefficients, where the lower indices sum to a constant, we can efficiently determine the value as a single binomial coefficient. The final calculated value is $924$.