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Bunuel
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I think it's E.
By method of counting - we can setup the following case: CVCVC, where C is a consonant and V is a vowel.

Since there are 21Cs and 5Vs in the English alphabet, the no. of possibilities for a CVCVC setup (wo repetition) is 21 x 5 x 20 x 4 x 19 = 159600

Now, there are 6 possible and valid arrangements in which we can place the alphabets - CVCVC, CVCCV, CCVCV, VCCCV, VCCVC, VCVCC

Hence, total possible arrangements is 159600 x 6 = 957600

Option E.
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Correct Answer: E (957,600)

GIVEN:
- 5-letter password: 2 vowels and 3 consonants from the alphabet (5 vowels, 21 consonants).
- All letters must be distinct.
- Vowels must NOT be adjacent.

STEP 1: SELECTION
- Vowel choices = 5C2 = 10
- Consonant choices = 21C3 = 1330

STEP 2: ARRANGEMENT (GAP METHOD)
- Arrange 3 consonants = 3! = 6
- Gaps created by consonants = 4 gaps (_ C1 _ C2 _ C3 _)
- Place 2 distinct vowels in 4 gaps = P(4, 2) = 12
- Total valid arrangements per set of 5 letters = 6 * 12 = 72

STEP 3: TOTAL COUNT
- Total Passwords = 10 * 1330 * 72 = 957,600

Conclusion: Answer Choice E (957,600).

Bunuel
A 5-letter password must consist of different letters of the alphabet, with 2 vowels and 3 consonants. If the vowels must not be adjacent to each other, how many such passwords can be created?

A. 95,760
B. 159.600
C. 256.000
D. 720,000
E. 957,600


Source: Math Revolution

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