The reasoning developed for counting microstates applies to many other situations involving probability. For example, if you flip a coin 5 times, how many different sequences of 3 heads and 2 tails are possible? Answer: 10 different sequences, such as HTHHT or TTHHH. In contrast, how many different sequences of 5 heads and 0 tails are possible? Obviously only one, HHHHH, and our equation gives , using the standard definition thatis defined to equal 1.

If the coin is equally likely on a single throw to come up heads or tails, any specific sequence like HTHHT or HHHHH is equally likely. However, there is only one way to get HHHHH, while there are 10 ways to get 3 heads and 2 tails, so this is 10 times more probable than getting all heads.

Use the expression to calculate the number of ways to get 0 heads, 1 head, 2 heads, 3 heads, 4 heads, or 5 heads in a sequence of 5 coin tosses. Make a graph of the number of ways vs. the number of heads.

The factorial is described as the product of the positive integers that range from 1 and go to 99. The sign of the factorial in the terms of n is defined as n!

Using the formula from the question, the number of the ways to get heads is expressed as:

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Here, is the number of ways to get head and is the number of heads.

$N=\frac{5!}{N\left(5-N\right)!}$

Substitute $N_0$$N\%MathType!MTEF!2!1!+-\%feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbbjxAHX\%garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wy\%Ubqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj-hEeeu0xXdbb\%a9frFj0-OqFfea0dXdd9vqaq-JfrVkFHe9pgea0dXdar-Jb9hs0dXd\%bPYxe9vr0-vr0-vqpWqaaeaabiGaciaacaqabeaadaabauaaaOabaq\%qabaGaamOtamaaBaaaleaacaaIWaaabeaakiabg2da9maalaaabaGa\%aGynaiaacgcaaeaacaaIWaGaaiyiamaabmaabaGaaGynaiabgkHiTi\%aaicdaaiaawIcacaGLPaaacaGGHaaaaaqaaiabg2da9maalaaabaGa\%aGynaiaacgcaaeaacaaIWaGaaiyiaiaaiwdacaGGHaaaaaqaaiabg2\%da9maalaaabaGaaGymaaqaaiaaicdacaGGHaaaaaqaaiabg2da9iaa\%igdaaaaa!4C61!\$\backslash begin\left\{gathered\right\}\left\{N\_0\right\}=\backslash frac\left\{\right\{5!\left\}\right\}\left\{\right\{0!\backslash left(\{5-0\}\backslash right)!\left\}\right\}\backslash \backslash =\backslash frac\left\{\right\{5!\left\}\right\}\left\{\right\{0!5!\left\}\right\}\backslash \backslash =\backslash frac\left\{1\right\}\left\{\right\{0!\left\}\right\}\backslash \backslash =1\backslash \backslash \backslash end\left\{gathered\right\}\$$uncaught exception: Invalid chunk

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$$\begin{gathered} N=\frac{5!}{0!\left(5-0\right)!} \\ =\frac{5!}{0!5!} \\ =\frac{1!}{0} \\ =1 \end{gathered}$$

The equation of the number of ways to get head is expressed as:

$N_1=\frac{N!}{N!\left(5-N\right)!}$

Here,$N_1$ is the number of ways to get head and is the number of heads.

Substitute the values in the above equation.

$$\begin{gathered} N_1=\frac{5!}{1!\left(5-1\right)!} \\ =\frac{5!}{1!4!} \\ =5 \end{gathered}$$

The equation of the number of ways to get head is expressed as:

$N_2=\frac{5!}{N!\left(5-N\right)!}$

Here, $N_2$is the number of ways to get head and is the number of heads.

Substitute the values in the above equation.

role="math" localid="1654854916246" $$\begin{gathered} N_2=\frac{5!}{2!3!} \\ =\frac{5.4}{2.1} \\ =10 \end{gathered}$$

The equation of the number of ways to get head is expressed as:

$N_3=\frac{5!}{N!\left(5-N\right)!}$

Here, $N_3$is the number of ways to get head and is the number of heads.

Substitute the values in the above equation.

$$\begin{gathered} N_3=\frac{5!}{3!\left(5-3\right)!} \\ =\frac{5!}{3!2!} \\ =\frac{5.4}{2.1} \\ =10 \end{gathered}$$

The equation of the number of ways to get head is expressed as:

$N_4=\frac{5!}{N!\left(5-N\right)!}$

Here, is the number of ways to get head and is the number of heads.

Substitute the values in the above equation.

$$\begin{gathered} N_4=\frac{5!}{N!\left(5-N\right)!} \\ =\frac{5!}{4!\left(5-4\right)!} \\ =\frac{5!}{4!1!} \\ =5 \end{gathered}$$

The equation of the number of ways to get head is expressed as:

$N_5=\frac{5!}{N!\left(5-N\right)!}$

Here, is the number of ways to get head and is the number of heads.

Substitute the values in the above equation.

role="math" localid="1654856485234" $$\begin{gathered} N_5=\frac{5!}{N!\left(5-N\right)!} \\ =\frac{5!}{5!\left(5-5\right)!} \\ =\frac{5!}{5!0!} \\ =1 \end{gathered}$$

Thus, the number of ways to get 0, 1, 2, 3, 4 and 5 heads in a sequence of 5 coin tosses are 1, 5, 10, 10, 1 and 0 respectively.