A 5-letter password must consist of different letters of the alphabet

Question

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

This is a  PS Butler Question

Check the links to other Butler Projects:

Data Sufficiency Butler  
  Critical Reasoning Butler

kevincan · Answer

Probably best to assume 21 consonants and 5 vowels

gautamK0206 · Answer

We start first by choosing the 2 Vowels and 3 Consonants from the pool of all the Vowels and Consonants.
5C2 = 10, 21C3 = 1330.
Now, to arrange the letters as per the constraints given, we first arrange the 3 Consonants, as 3! = 6
This would look something like this: _ C _ C _ C _
This leaves us 4 spots to place the vowels, that can be done by 4P2 = 12,
So, the final solution would look like: 10*1330*6*12
The correct answer is  E. 957,600

abhikc3004 · Answer

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.

counterclash1 · Answer

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).

A 5-letter password must consist of different letters of the alphabet

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

GMAT Problem Solving (PS) Questions

A 5-letter password must consist of different letters of the alphabet

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia