How many 3-letter code words can be formed if at least one of the lett

Question

How many 3-letter code words can be formed from the letters of the English alphabet, if at least one of the letters is to be chosen from the vowels a, e, i, o, and u?

A. 3,380
B. 6,615
C. 8,315
D. 10,140
E. 12,215

ShaikhMoice · Accepted Answer

Total ways without restriction= 26*26*26
Total ways with only consonants= 21*21*21
Total ways with atleast 1 Vowel= 26^3-21^3

MHIKER · Answer

The problem didn't tell us that we can't repeat. And we have to confirm at least one vowel, so we have more than one vowel too.

For the first position, we can take one from the 26 alphabet

For the second position, we can take from the 26 alphabets again

For the third position, we will confirm the vowel from the 5, because if we don't get any vowel in the first try. We don't need to maintain any order.

Thus the ways:  26*26*5=3,380

The answer is A

MHIKER · Answer

I answered A but the OA is C. Would you please help me to learn. I will be obliged.

MM2008 · Answer

Why do we not get the same result if we do (21*21*5) + (21*5*5) + (5*5*5)
i.e., (2 consonants and 1 vowel OR 1 consonant and 2 vowels OR 3 vowels)?

Bunuel · Answer

That approach does not account for the different permutations of the first two cases, so it should be 3*(21*21*5) + 3*(21*5*5) + (5*5*5)

AninT · Answer

Why is 5*26*26*3! wrong? I considered 3! for the arrangement of the letters selected.

Bunuel · Answer

If one or more letters are the same, the number of arrangements wouldn’t be 3!. For example, if you get the case "a, a, a," the number of permutations would be just 1, not 3!. So, considering 3! for all cases leads to an overcount.

anishd09 · Answer

My approach-

We have 3 places to fill and at least 1 vowel

So we can have 3 cases-

Case-1 1 Vowel and 2 constants 5*21*21 and the permutation for this case will be 3!/2!
 5*21*21*3=6615

Case-2 2 Vowel and 1 constants 5*5*21 and the permutation for this case will be 3!/2!
 5*5*21*3=1575

Case-3 3 Vowels 5*5*5=125

6615+1517+125=8315

The answer is option C

Kinshook · Answer

How many 3-letter code words can be formed from the letters of the English alphabet, if at least one of the letters is to be chosen from the vowels a, e, i, o, and u?

3-letter words = 26^3=17,576 
3-letter words without any vowel = 21^3=9,261

3-letter words with at least 1 vowel = 26^3-21^3 =8,315

IMO C

ravi1522 · Answer

How many 3-letter code words can be formed from the letters of the English alphabet, if at least one of the letters is to be chosen from the vowels a, e, i, o, and u?

total 3 letter word =26*26*26
total 3 letter word without vowel =21*21*21 
total letter with at least vowel =26*26*26-21*21*21=8315

Hence option C is correct

How many 3-letter code words can be formed if at least one of the lett

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GMAT Problem Solving (PS) Questions

How many 3-letter code words can be formed if at least one of the lett

Prep Toolkit

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My Rewards