I have a statement thay says:
If they are formed, 5-letter words (with or without meaning) with the letters of the word PROBLEMA, So which one (s) of the following statements is (are) true (s)?
I) 120 words contains only consonants.
II) 240 words have E and A at the extremes.
III) 7! words starts with L
Then, my development was:
$I)$ True, because taking into consideration only consonants is equal to $5!=120.$
$II)$ True, because if I have $E\ \_\ \_\ \_\ A$ and, $A\ \_\ \_\ \_\ E$, so I have two cases, and for each case must take $3$ letters, of $6$, importing the order, since $ABC$ will be different than $BAC$, then it is a variation, whose formula is $$\frac{n!}{(n−r)!},$$
so $\Large\frac{6!}{(6−3)!}$ $\cdot 2!=120\cdot 2!=240.$
$III)$ True, because if the word start with $${\large L\ \_\ \_\ \_\ \_\ \_}\ ,$$
I have $7!$ of ordering the other elements, since it is a circular permutation, because $L$ is fixed.
My problem is that the correct answer must be only I) and II), I would like to know, why the III) is incorrect and also if there is a faster procedure for my answers I) and II)
III is false. $\dfrac{7!}{3!}$ words start with an L. There are only five letters in the generated word. $7!$ would imply there are 8 letters in this word.
And your methodology for figuring out I and II seemed appropriate to me. I cannot think of a "faster" method.