What is the coefficient of the $x^6 y^3$ term in the expansion of $(x+2 y)^9$ ?
The objective is to find the coefficient of the term \(x^6y^3\) in the expansion of the binomial expression \((x+2y)^9\). The binomial theorem states that the expansion of \((a+b)^n\) is given by \(\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). For our specific expression \((x+2y)^9\), we have \(a = x\), \(b = 2y\), and \(n = 9\).
The general term \(T_{k+1}\) in the expansion of \((x+2y)^9\) is given by:
\[T_{k+1} = \binom{9}{k} x^{9-k} (2y)^k\]
By simplifying this, we get:
\[T_{k+1} = \binom{9}{k} 2^k x^{9-k} y^k\]
We are interested in the term involving \(x^6y^3\). Comparing the powers of \(x\) and \(y\) in the general term to those in \(x^6y^3\), we set the exponent of \(y\) equal to \(3\), which gives \(k = 3\). Substituting \(k=3\) into the exponent of \(x\), we get \(9-k = 9-3 = 6\), which matches the exponent of \(x\) in our target term.
Now, we substitute \(k=3\) into the expression for the general term:
\[T_{3+1} = \binom{9}{3} 2^3 x^{9-3} y^3 = \binom{9}{3} \times 8 \times x^6 y^3\]
First, calculate the binomial coefficient \(\binom{9}{3}\):
\[\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84\]
Next, multiply this by the power of \(2\):
\[84 \times 2^3 = 84 \times 8 = 672\]
Thus, the coefficient of the \(x^6 y^3\) term is \(672\).
The binomial theorem is widely used in probability theory and statistics, such as calculating the number of possible outcomes when flipping a coin or rolling dice multiple times. For example, if you want to determine the probability of getting exactly 3 heads in 9 coin flips, you use the binomial coefficient \(\binom{9}{3}\) to find the number of ways to choose which 3 flips result in heads. Mastering this expansion technique allows you to quickly find specific probabilities or solve complex combinatorics problems in real-world simulations and predictive modeling.