Question
If all words with 2 distinct vowels and 3 distinct consonants were listed alphabetically, what would be the rank of “ACDEF’?
A. 4716
B. 4720
C. 4718
D. 4717
Correct Answer
Choice C
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 ^{20}C_{2}and ^{4}C_{1} ways. Then the three distinct letters can be rearranged in 3! Ways.
Or, number of words starting with AB = ^{20}C_{2}* ^{4}C_{1 }* 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. = ^{19}C_{1} * ^{4}C_{1}. This can be rearranged in 2! Ways.
Or, number of words starting with ACB = ^{19}C_{1} * ^{4}C_{1}* 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)