A 3-letters code consists of three different letters. If the code

Question

A 3-letters code consists of three different letters. If the code contains at least 2 vowels, how many such codes are possible?

A. 220
B. 480
C. 660
D. 1320
E. 2880

jeeteshsingh · Answer

VVC + VVV

where V = Vowel
C = Consonant

VVC = 5c1 x 4c1 x 21c1 x 3! = 5x4x21x6= 2520

VVV = 5c1 x 4c1 x 3c1 x 3! = 5 x 4 x 3 x 6 = 360

VVC + VVV = 2880 (Ans)

chix475ntu · Answer

2 vowels and 1 consonant - can arrange in 3 ways VVC, VCV, CVV = 5 * 4 * 21 * 3 = 1260
3 vowels - VVV - 5 * 4 * 3 = 60

total = 1320

jeeteshsingh · Answer

Its a code we are talking abt. Hence the arrangement between two vowels would also matter...

as per you have considered only VVC.... that means ... AEX ... and EAX is the same.... which is wrong! Please check.

chix475ntu · Answer

we are already counting that in (5 * 4) ... for example ..
how many ways to select 2 out of a e i o u where order doesnt matter - 5 * 4 / 2 = 10 and the options are 
ae, ai, ao, au, ei, eo, eu, io, iu, ou ... COMBINATIONS
if order matters than each one will have its counterpart .. example ae, ea .. and the number becomes 5 * 4 = 20 - PERMUTATIONS ..

you are taking permutations and then multipluing those again with 3! to get duplicates.

honeyrai · Answer

Answer is: (5C2. 21C1. 3!) + (5P3) or (5C2. 21C1. 3!) + (5C3 * 3!) == 1260 + 60= 1320.

manpreetsingh86 · Answer

following 2 cases are possible.

case 1: code contains 2 vowels and 1 consonant= 5C2.21C1 (5= total no. of vowels, 21= total no. of consonants)

case 2: code contains 3 vowels. = 5C3

thus total number of codes = 5C2.21C1 + 5C3= 210+10 = 220

desaichinmay22 · Answer

Hi,

You will have to multiply 220 by 3!. You have taken into account only selection of letters and not their arrangement.

manpreetsingh86 · Answer

yes, i forgot to multiply 220 with 3!.

Bunuel · Answer

Merging similar topics. Please refer to the discussion above and ask if anything remains unclear. Check Constructing Numbers, Codes and Passwords in our Special Questions Directory . P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 1, and 3. Thank you.

TooLong150 · Answer

5 Vowels
21 Consonants

Remember that the letters have to  different .

At least 2 = 2-Vowels + 3-Vowels
2-Vowels: 5*4*21 = 420
3-Vowels: 5*4*3 = 60

Total = 480

energetics · Answer

We have 5 vowels (A,E,I,O,U) and 21 consonants.

Restriction 1: must have 3 different letters
Restriction 2: must have at least 2 vowels

Case 1: 3 vowels = 5*4*3 = 60 possibilities
Case 2: 2 vowels, 1 consonant = 5*4*21 *3(for the position of the consonant)

Add together: 60+1260 = 1320

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A 3-letters code consists of three different letters. If the code

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