Question

${}^{10}C_0 {}^{20}C_{10} - {}^{10}C_1 {}^{18}C_{10} + {}^{10}C_2 {}^{16}C_{10} - {}^{10}C_3 {}^{14}C_{10} + {}^{10}C_4 {}^{12}C_{10} - {}^{10}C_5 {}^{10}C_{10}$ is equal to

Answer 1

1024

Answer 2

The given expression is: $S = {}^{10}C_0 {}^{20}C_{10} - {}^{10}C_1 {}^{18}C_{10} + {}^{10}C_2 {}^{16}C_{10} - {}^{10}C_3 {}^{14}C_{10} + {}^{10}C_4 {}^{12}C_{10} - {}^{10}C_5 {}^{10}C_{10}$

We can write this sum in a more compact form using summation notation. The general term is $(-1)^k {}^{10}C_k {}^{20-2k}C_{10}$. So, $S = \sum_{k=0}^{5} (-1)^k {}^{10}C_k {}^{20-2k}C_{10}$.

Observe the term ${}^{20-2k}C_{10}$. If $20-2k < 10$, then this binomial coefficient is zero. This happens when $10 < 2k$, or $k > 5$. For example: If $k=6$, the term is ${}^{20-12}C_{10} = {}^{8}C_{10} = 0$. If $k=7$, the term is ${}^{20-14}C_{10} = {}^{6}C_{10} = 0$. ... If $k=10$, the term is ${}^{20-20}C_{10} = {}^{0}C_{10} = 0$.

Since all terms from $k=6$ to $k=10$ are zero, we can extend the upper limit of the summation from 5 to 10 without changing the value of the sum: $S = \sum_{k=0}^{10} (-1)^k {}^{10}C_k {}^{20-2k}C_{10}$.

Now, we use the property that ${}^{n}C_r$ is the coefficient of $x^r$ in the expansion of $(1+x)^n$. So, ${}^{20-2k}C_{10}$ is the coefficient of $x^{10}$ in $(1+x)^{20-2k}$. Therefore, we can write $S$ as: $S = \sum_{k=0}^{10} (-1)^k {}^{10}C_k \left[ \text{Coeff of } x^{10} \text{ in } (1+x)^{20-2k} \right]$ Since the coefficient extraction is a linear operation, we can move it outside the summation: $S = \text{Coeff of } x^{10} \text{ in } \left[ \sum_{k=0}^{10} (-1)^k {}^{10}C_k (1+x)^{20-2k} \right]$

Factor out $(1+x)^{20}$ from each term inside the summation: $S = \text{Coeff of } x^{10} \text{ in } \left[ (1+x)^{20} \sum_{k=0}^{10} (-1)^k {}^{10}C_k (1+x)^{-2k} \right]$ The summation part $\sum_{k=0}^{10} (-1)^k {}^{10}C_k (1+x)^{-2k}$ is a binomial expansion of the form $\sum_{k=0}^{n} {}^{n}C_k (-y)^k = (1-y)^n$. Here, $n=10$ and $y=(1+x)^{-2}$. So, $\sum_{k=0}^{10} (-1)^k {}^{10}C_k (1+x)^{-2k} = \left(1 - (1+x)^{-2}\right)^{10}$.

Substitute this back into the expression for $S$: $S = \text{Coeff of } x^{10} \text{ in } \left[ (1+x)^{20} \left(1 - \frac{1}{(1+x)^2}\right)^{10} \right]$ Simplify the term inside the parenthesis: $1 - \frac{1}{(1+x)^2} = \frac{(1+x)^2 - 1}{(1+x)^2} = \frac{(1+2x+x^2) - 1}{(1+x)^2} = \frac{2x+x^2}{(1+x)^2}$

Now substitute this back: $S = \text{Coeff of } x^{10} \text{ in } \left[ (1+x)^{20} \left(\frac{2x+x^2}{(1+x)^2}\right)^{10} \right]$ $S = \text{Coeff of } x^{10} \text{ in } \left[ (1+x)^{20} \frac{(2x+x^2)^{10}}{((1+x)^2)^{10}} \right]$ $S = \text{Coeff of } x^{10} \text{ in } \left[ (1+x)^{20} \frac{(2x+x^2)^{10}}{(1+x)^{20}} \right]$ The $(1+x)^{20}$ terms cancel out: $S = \text{Coeff of } x^{10} \text{ in } (2x+x^2)^{10}$

Factor out $x$ from $(2x+x^2)$: $2x+x^2 = x(2+x)$ So, $(2x+x^2)^{10} = (x(2+x))^{10} = x^{10}(2+x)^{10}$.

Therefore, $S = \text{Coeff of } x^{10} \text{ in } x^{10}(2+x)^{10}$. To find the coefficient of $x^{10}$ in $x^{10}(2+x)^{10}$, we need to find the constant term (coefficient of $x^0$) in $(2+x)^{10}$. The expansion of $(2+x)^{10}$ is given by the binomial theorem: $(2+x)^{10} = \sum_{r=0}^{10} {}^{10}C_r 2^{10-r} x^r$. The constant term (when $r=0$) is ${}^{10}C_0 2^{10-0} x^0 = 1 \cdot 2^{10} \cdot 1 = 2^{10}$.

So, $S = 2^{10}$. $2^{10} = 1024$.