Question

If \(C_0, C_1, C_2, ..., C_{10}\) represent the binomial coefficients in the expansion of \((1+x)^{10}\), then \(C_0C_6 + C_1C_7 + C_2C_8 + C_3C_9 + C_4C_{10} = \)

• A. 9690
• B. 4845
• C. 1615
• D. 3230

Hint:

The sum \(\sum C_r C_k\) often relates to the coefficient of a term in the product of two binomial expansions. Using the identity \(^nC_r = ^nC_{n-r}\) is key to transforming the sum into a recognizable form, typically the coefficient of \(x^k\) in \((1+x)^{2n}\), which is \(^{2n}C_k\).

Answer 1

The Correct Option is B

We need to find the sum \(S = C_0C_6 + C_1C_7 + C_2C_8 + C_3C_9 + C_4C_{10}\), where \(C_r = ^{10}C_r\).
We use the binomial coefficient identity \(^nC_r = ^nC_{n-r}\). For \(n=10\), we have \(C_r = C_{10-r}\).
Let's apply this identity to the second factor in each term of the sum.
\(C_6 = ^{10}C_6 = ^{10}C_{10-6} = ^{10}C_4 = C_4\).
\(C_7 = ^{10}C_7 = ^{10}C_{10-7} = ^{10}C_3 = C_3\).
\(C_8 = ^{10}C_8 = ^{10}C_{10-8} = ^{10}C_2 = C_2\).
\(C_9 = ^{10}C_9 = ^{10}C_{10-9} = ^{10}C_1 = C_1\).
\(C_{10} = ^{10}C_{10} = ^{10}C_{10-10} = ^{10}C_0 = C_0\).
Substituting these into the sum:
\(S = C_0C_4 + C_1C_3 + C_2C_2 + C_3C_1 + C_4C_0\).
This sum is a specific case of Vandermonde's Identity, which states that the coefficient of \(x^k\) in the expansion of \((1+x)^{m+n}\) is \(\sum_{r=0}^k (^mC_r)(^nC_{k-r})\).
Our sum \(S = \sum_{r=0}^4 C_r C_{4-r}\) matches this form. It represents the coefficient of \(x^4\) in the product of two expansions.
The sum is the coefficient of \(x^4\) in the expansion of \((C_0+C_1x+C_2x^2+...)(C_0+C_1x+C_2x^2+...)\).
This product is \((1+x)^{10} \times (1+x)^{10} = (1+x)^{20}\).
So, we need to find the coefficient of \(x^4\) in the expansion of \((1+x)^{20}\).
This coefficient is \(^{20}C_4\).
\(^{20}C_4 = \frac{20!}{4!(20-4)!} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1}\).
\(^{20}C_4 = (5 \times 19 \times 3 \times 17) = 4845\).