joepc wrote:

shashankism wrote:

joepc wrote:

Dear Math Experts,

What will be answer if repetition in CVCVC is not allowed(once a vowel or consonant is used it cannot be used again)

Thanks

Joepc

In that case 1st C can take 3 values, 2nd C can take 2 values and 3rd C can take 1 value.

Similarly 1st V can take 2 values and 2nd V van take 1 value..

So, total no. of words formed = 3*2*1*2*1 = 12...

Thanks for your reply Shashankism, Is it not 120?, below is my reasoning Please correct me where I am wrong

IT is CVCVC

Considering Consonants3 Consonants 1st C in 3 ways, 2nd C in 2 Ways and Third C in 1 Ways = 3*2*1 = 6 ways

Considering VowelsOut of 5 vowels 2 vowels can be selected in 5C2 = 10 ways and it(2 vowels) can be permuted 2! ways which 20

Hence it is 6*20 = 120 ways

But the question stem says there are only 2 vowels(check question it says there are 2 unique vowels misspelled as values and 3 unique consonant)..

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