How many different ways are there to get 5 heads in 10 throws of a true coin? How many different ways are there to get no heads in 10 throws of a true coin?

The given data can be listed below as,

When a coin tossed 10 times, then each coin will have 2 possible outcomes. So, the total number of outcomes can be expressed as follows:

Substitute all the known values in the above equation.

role="math" localid="1655703002075" $$\begin{gathered} \mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}=2^{10} \\ =1024 \end{gathered}$$

The expression to calculate the number of different ways to get exactly 5 heads in 10 throws is expressed as follows:.$\mathrm{Number}\mathrm{of}\mathrm{different}\mathrm{ways}=\frac{\mathrm{n}!}{5!\times 5!}$

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{different}\mathrm{ways}=\frac{10!}{5!\times 5!} \\ =\left(\frac{6\times 7\times 8\times 9\times 10}{1\times 2\times 3\times 4\times 5}\right) \\ =252 \end{gathered}$$

Thus, the number of different ways to get exactly 5 heads in 10 throws is 252.

The number of ways to get no heads in 10 throws of a true coin is only one that is in every 10 throws only tails appears on the true coin.

Thus, the number of ways to get no heads in 10 throws of a true coin is 1.