Probability
So I have been trying to solve this question in probability, but I don't seem to get the correct answer. I am not bad at probability and this seems to be easy one, but I'm just struggling really bad on this one. So the question is this:
What's the probability of choosing $5$ letters from $a, b, g, d, e, z$ (these are the first $6$ letters in Armenian alphabet) so that the word "bad" will be included in it.
P.S. the letters cannot be repeated and "bad" can be anywhere in a $5$-letter word.
I know the answer, but I can't get it. I am trying to solve this in this way:
$$P = \frac{3!}{A(6, 5)}$$
$\color{blue}{A(n, k) = C(n, k) \cdot k!}$
Would be thankful if you could help me out.
Here's the answer below for people wondering ↓
$P = \frac{1}{40}$
To calculate the probability of the event
$A$= you have the word "BAD" in the chosen letter combination
we need to evaluate
$$ P(A)=\frac{\text{A happened}}{\text{all possible outcomes}}. $$
All possible outcomes are $\binom{6}{5}\cdot 5!=6\cdot 5!$ (You can think about this as we remove one letter and pick the rest hence the $6$ different options and we can arrange these in $5!$ different ways). To find the number of outcomes when we have the word "BAD" we will glue these letters together as one entity, let us refer to it as $\sigma$ from now on. This will of course always be chosen and then we have $3$ letters left to choose from and we need to fill $2$ empty places in the constructed word. You can choose the remaining two letters in $\binom{3}{2}$ different ways. Now we have $\sigma$ and $2$ more letters and these can be arranged in $3!$ different ways. So altogether we have $$ \binom{3}{2}\cdot 3! $$ different ways to construct a desired word.
Hope I could help.