The question is from Permutation and Combination. It is about combination of letter which is ultimately words. We need to find out the rank of a particular five letter word. This section hosts a number of questions which are on par with CAT questions in difficulty on CAT Permutation and Combination, and CAT Probability.

Question 6: If all words with 2 distinct vowels and 3 distinct consonants were listed alphabetically, what would be the rank of “ACDEF’?

- 4716
- 4720
- 4718
- 1717

4718

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

The question is **"what would be the rank of “ACDEF’?"**

Choice C is the correct answer.

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