CAT Quantitative Aptitude Questions | CAT Permutation and Combination Questions

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’?

1. 4716
2. 4720
3. 4718
4. 1717

Explanatory Answer

Method of solving this CAT Question from Permutation and Combination: Thankfully we do not need to worry about words that make sense.

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

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

Hence the answer is "4718"

Choice C is the correct answer.