The number of arrangements that can be formed from the letters $$a, b, c, d, e, f$$ taken $$3$$ at a time without repetition and each arrangement containing at least one vowel, is

Total number of 3-letter arrangements that can be made from the 6 distinct letters $$a, b, c, d, e, f$$ (no restriction) is the number of permutations of 6 objects taken 3 at a time.

Using $$\,{}^{n}P_{r} = n(n-1)\dots (n-r+1)\,,$$
we get $$\,{}^{6}P_{3} = 6 \times 5 \times 4 = 120.$$

Now subtract the arrangements that contain no vowel. The vowels present are $$a$$ and $$e,$$ so the consonants are $$b, c, d, f.$$

Select and arrange 3 letters from these 4 consonants:
$$\,{}^{4}P_{3} = 4 \times 3 \times 2 = 24.$$

Arrangements having at least one vowel = (all arrangements) − (arrangements with no vowel)
$$120 - 24 = 96.$$

Option A which is: $$96$$