what is the coefficient of the $x^{5}y^{5}$ - term in the binomial expansion of $(2x - 3y)^{10}$?\n$_{10}c_{5…

Explanation:

Step1: Recall binomial theorem formula

The binomial theorem states that $(a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}$. In the expansion of $(2x-3y)^{10}$, we have $a = 2x$, $b=- 3y$ and $n = 10$.

Step2: Find the value of k for the $x^{5}y^{5}$ - term

We want to find the coefficient of the $x^{5}y^{5}$ - term. Since $a = 2x$ and $b=-3y$, and $a^{n - k}b^{k}=(2x)^{n - k}(-3y)^{k}$, we set $n - k=5$ and $k = 5$ (because for $(2x)^{n - k}(-3y)^{k}=2^{n - k}x^{n - k}(-3)^{k}y^{k}$, when $n - k = 5$ and $k = 5$, we get $x^{5}y^{5}$). Here $n=10$.

Step3: Calculate the coefficient

The general term of the binomial expansion is $T_{k + 1}=\binom{n}{k}a^{n - k}b^{k}$. Substituting $n = 10$, $k = 5$, $a = 2x$ and $b=-3y$ into the formula, we get $T_{6}=\binom{10}{5}(2x)^{10 - 5}(-3y)^{5}=\binom{10}{5}(2)^{5}x^{5}(-3)^{5}y^{5}$. So the coefficient of the $x^{5}y^{5}$ - term is $\binom{10}{5}(2)^{5}(-3)^{5}$.

Answer:

B. $\binom{10}{5}(2)^{5}(-3)^{5}$