Question

If all words with 2 distinct vowels and 3 distinct consonants were listed alphabetically, what would be the rank of “ACDEF’?

1. 4716
2. 4720
3. 4718
4. 4717

Explanatory Answer

The first word would be ABCDE. With 2 distinct vowels, 3 distinct consonants, this is the first word we can come up with.

Starting with AB, we can have a number of words.

AB __ __ __. The next three slots should have 2 consonants and one vowel. This can be selected in ²⁰C₂ and ⁴C₁ ways. Then the three distinct letters can be rearranged in 3! Ways.

Or, number of words starting with AB = ²⁰C₂ * ⁴C₁ * 3! = 190 * 4 * 6 = 4560

Next, we move on to words starting with ACB
ACB __ __. The last two slots have to be filled with one vowel and one consonant. = ¹⁹C₁ * ⁴C₁. This can be rearranged in 2! Ways.

Or, number of words starting with ACB = ¹⁹C₁ * ⁴C₁ * 2 = 19 * 4 * 2 = 152

Next we move on words starting with ACDB: There are 4 different words on this list – ACDBE, ACDBI, ACDBO, ACDBU

So, far number of words gone = 4560 + 152 + 4 = 4716

Starting with AB          4560
Starting with ACB         152
Starting with ACDB          4
Total words gone        4716

After this we move to words starting with ACDE, the first possible word is ACDEB. After this we have ACDEF.

So, rank of ACDEF = 4718

Answer Choice (C)