When a coin is tossed n times and the proabability for getting 6 heads is equal to the probability of getting 8 heads, then the value of n is 
 

when tossing a coin, 

Probability of getting head = Probability of getting tail $=\frac{1}{2}$

$\Rightarrow \mathrm{p}=\frac{1}{2}, \mathrm{q}=\frac{1}{2}$ where 'X' follows binomial distribution.

Given that $\mathrm{P}(\mathrm{x}=6)=\mathrm{P}(\mathrm{x}=8)$

$$\begin{gathered} \Rightarrow {}^{\mathrm{n}}{\mathrm{C}}_6 {\mathrm{p}}^6 {\left(\mathrm{q}\right)}^{\mathrm{n}-6}={}^{\mathrm{n}}{\mathrm{C}}_8 {\left(\mathrm{p}\right)}^8 {\left(\mathrm{q}\right)}^{\mathrm{n}-8} \\ \Rightarrow {}^{\mathrm{n}}{\mathrm{C}}_6 {\left(\frac{1}{2}\right)}^{\mathrm{n}}={}^{\mathrm{n}}{\mathrm{C}}_8 {\left(\frac{1}{2}\right)}^{\mathrm{n}} \\ \Rightarrow {}^{\mathrm{n}}{\mathrm{C}}_6={}^{\mathrm{n}}{\mathrm{C}}_8 \\ \Rightarrow \mathrm{n}=6+8=14 \left(\because {}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}={}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{s}} \Rightarrow \mathrm{n}=\mathrm{r}+\mathrm{s}\right) \end{gathered}$$