**Question:**

Find the sixth term in the expansion $\left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)^{n}$, if the binomial coefficient of the third term from the end is 45 .

**Solution:**

In the binomial expansion of $\left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)^{n}$, there are $(n+1)$ terms.

The third term from the end in the expansion of $\left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)^{n}$, is the third term from the beginning in the expansion of $\left(x^{\frac{1}{3}}+y^{\frac{1}{2}}\right)^{n}$.

$\therefore$ The binomial coefficient of the third term from the end $={ }^{n} C_{2}$

If is given that the binomial coefficient of the third term from the end is 45.

$\therefore{ }^{n} C_{2}=45$

$\Rightarrow \frac{n(n-1)}{2}=45$

$\Rightarrow n^{2}-n-90=0$

$\Rightarrow(n-10)(n+9)=0$

$\Rightarrow n=10 \quad(\because n$ cannot be negative $)$

Let $T_{6}$ be the sixth term in the binomial expansion of $\left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)^{n}$. Then

$T_{6}={ }^{n} C_{5}\left(y^{\frac{1}{2}}\right)^{n-5}\left(x^{\frac{1}{3}}\right)^{5}=^{10} C_{5} y^{\frac{5}{2}} x^{\frac{5}{3}}=252 y^{\frac{5}{2}} x^{\frac{5}{3}}$

Hence, the sixth term in the expansion of $\left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)^{n}$, is $252 y^{\frac{5}{2}} x^{\frac{5}{3}}$.