Question

# Consider n independent sequential trials, each of which is successful with probability p. If there is a total of k successes, show that each of then!/[k!(n−k)!]possible arrangements of the k successes and n-k failures is equally likely.

Probability
Consider n independent sequential trials, each of which is successful with probability p. If there is a total of k successes, show that each of the
$$\displaystyle{n}\frac{!}{{{k}!{\left({n}−{k}\right)}!}}$$
possible arrangements of the k successes and n-k failures is equally likely.

2021-08-16
Probability that exactly one of the event E or F occurs is
$$\displaystyle{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}$$
Now,
$$\displaystyle{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}={P}{\left({E}{F}^{{c}}\right)}+{P}{\left({E}^{{c}}{F}\right)}-{P}{\left({E}{F}^{{c}}\cap{E}^{{c}}{F}\right)}$$
Since $$\displaystyle{P}{\left({E}{F}^{{c}}\cap{E}^{{c}}{F}\right)}={0}$$
$$\displaystyle\Rightarrow{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}={P}{\left({E}{F}^{{c}}\right)}+{P}{\left({E}^{{c}}{F}\right)}$$
Also, $$\displaystyle{P}{\left({E}{F}^{{c}}\right)}={P}{\left({E}\right)}-{P}{\left({E}{F}\right)}$$ and $$\displaystyle{P}{\left({E}^{{c}}{F}\right)}={P}{\left({F}\right)}-{P}{\left({E}{F}\right)}\Rightarrow{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}={P}{\left({E}\right)}+{P}{\left({F}\right)}-{2}{P}{\left({E}{F}\right)}$$
Hence proved!