# CAT Counting: Letters and Words

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

Choice C

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

IIM CAT Preparation Tips: CAT Counting: Letters and Words

## May 31, 2013

### CAT Counting: Letters and Words

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

Choice C

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