18) In how many distinct ways can the letters $\mathrm{I}, \mathrm{S}, \mathrm{O}, \mathrm{L}, \mathrm{A}$ be arranged so that no two consonants are consecutive?
(A) 60,
(B) 72,
(C) 84,
(D) 96
(E) 120.
18. The answer is $\mathbf{( B )}$.

It is convenient to calculate the required number by the difference between the total number of permutations (i.e., $5 !=120$) and those in which we find two consecutive consonants.

There are four distinct ways to have two consecutive consonants: at positions 1 and 2, at positions 2 and 3, 3 and 4, 4 and 5. For each of these possibilities, we can arrange the two consonants in two different ways (SL and LS) and the three vowels in $3 !=6$ ways. Consequently, the total number of arrangements of the letters of ISOLA in such a way that two consonants are consecutive is $4 \cdot 2 \cdot 6=48$. The required number of arrangements is then $120-48=72$.