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
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)
Labels: CAT Counting, CAT letters and words, CAT Permutation Combination
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