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Each letter of a 5-letter password must be a different letter of the a

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Math Revolution GMAT Instructor
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Each letter of a 5-letter password must be a different letter of the a  [#permalink]

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New post 11 Apr 2018, 05:58
1
4
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A
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D
E

Difficulty:

  95% (hard)

Question Stats:

23% (03:19) correct 77% (02:46) wrong based on 43 sessions

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[GMAT math practice question]

Each letter of a 5-letter password must be a different letter of the alphabet. Each password must contain 2 vowels and 3 consonants, and the vowels must not be adjacent to each other. How many different passwords of this form can be made?

A. 957600
B. 128000
C. 159600
D. 256000
E. 720000

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Re: Each letter of a 5-letter password must be a different letter of the a  [#permalink]

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New post 11 Apr 2018, 09:27
2
2
MathRevolution wrote:
[GMAT math practice question]

Each letter of a 5-letter password must be a different letter of the alphabet. Each password must contain 2 vowels and 3 consonants, and the vowels must not be adjacent to each other. How many different passwords of this form can be made?

A. 95760
B. 128000
C. 159600
D. 256000
E. 720000


Place the consonants in three places with a gap between each.. ways = 21P3 = 21*20*19
choose any two gaps from 4 gaps formed..... _C_C_C_ ------- 4C2=6
Place 2 vowels there = 5P2= 20..

answer = 21*20*19*6*20 = 957600
A with extra 0
editing
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Each letter of a 5-letter password must be a different letter of the a  [#permalink]

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New post Updated on: 15 Apr 2018, 20:33
1
1
=>

The total number of passwords with 2 vowels and 3 consonants is determined as follows:

1) Choose the 2 different vowels in 5C2 ways
2) Choose the 3 consonants in 21C3 ways
3) Arrange the 5 letters in 5! ways
So, the number of passwords is
21C3 * 5C2 * 5! = ( 21*20*19 ) / (1*2*3) * (5*4)/(1*2) * 120 = 1330 * 10 * 120 = 1596000.

The number of passwords with 2 adjacent vowels and 3 consonants is
determined as follows:

1) Choose the 2 different vowels in 5C2 ways
2) Choose the 3 consonants in 21C3 ways
3) Choose a position for the two adjacent vowels in 4! ways
4) Rearrange the adjacent vowels in 2! ways
So, the number of passwords with two adjacent vowels is

21C3 * 5C2 * 4!*2! = ( 21*20*19 ) / (1*2*3) * (5*4)/(1*2) * 48 = 1330 * 10 * 48 = 638400.

Therefore, the number of passwords without two adjacent vowels is
1596000 – 638400 = 957600.

Therefore, the answer is A.

Answer: A
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Originally posted by MathRevolution on 13 Apr 2018, 02:40.
Last edited by MathRevolution on 15 Apr 2018, 20:33, edited 2 times in total.
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Re: Each letter of a 5-letter password must be a different letter of the a  [#permalink]

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New post 13 Apr 2018, 02:54
1
MathRevolution wrote:
=>

The total number of passwords with 2 vowels and 3 consonants is determined as follows:

1) Choose the 2 different vowels in 5C2 ways
2) Choose the 3 consonants in 21C3 ways
3) Arrange the 5 letters in 5! ways
So, the number of passwords is
21C3 * 5C2 * 5! = ( 21*20*19 ) / (1*2*3) * (5*4)/(1*2) * 120 = 133 * 10 * 120 = 159600.

The number of passwords with 2 adjacent vowels and 3 consonants is
determined as follows:

1) Choose the 2 different vowels in 5C2 ways
2) Choose the 3 consonants in 21C3 ways
3) Choose a position for the two adjacent vowels in 4! ways
4) Rearrange the adjacent vowels in 2! ways
So, the number of passwords with two adjacent vowels is

21C3 * 5C2 * 4!*2! = ( 21*20*19 ) / (1*2*3) * (5*4)/(1*2) * 48 = 133 * 10 * 48 = 63840.

Therefore, the number of passwords without two adjacent vowels is
159600 – 63840 = 95760.

Therefore, the answer is A.

Answer: A


hi..

you are wrong in the coloured portion....
21*20*19/6 is 1330 whereas you have taken it as 133..

therefore the answer you have got is to be multiplied with 10.. ans is 957600 and NOT 95760
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Each letter of a 5-letter password must be a different letter of the a  [#permalink]

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New post 13 Apr 2018, 17:56
1
21 = consonants
5 = vowels

21*20*19*5*4 = 159600

Now there are 6 different ways to organize the password

VCVCC
VCCVC
VCCCV
CVCVC
CVCCV
CCVCV

159600*6 = 957600
Answer A
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Re: Each letter of a 5-letter password must be a different letter of the a  [#permalink]

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New post 22 Apr 2018, 05:38
5 vowel and 21 consonants

5C2 *21C3*(5!-(4!*2))

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Re: Each letter of a 5-letter password must be a different letter of the a &nbs [#permalink] 22 Apr 2018, 05:38
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