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Arranging the letters

Expert replies
by talaangoshtari » Fri Sep 25, 2015 9:16 pm
In how many ways can letters of the word NUMBERS be arranged so that vowels are in alternate position?

A. 576
B. 1172
C. 1200
D. 1500
E. 1800
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Source: — Problem Solving |

by adthedaddy » Fri Sep 25, 2015 11:13 pm
This is a mixed Permutation and Combination problem.

There are 2 vowels in the word 'NUMBERS'.

There are 5 possible positions of the seven letter word NUMBERS where the vowels can be arranged alternately.
Using V for vowel and C for consonant, they are -
VCVCCCC
CVCVCCC
CCVCVCC
CCCVCVC
CCCCVCV

Thus, 2 vowels can be placed in the given 5 possibilities in 5C2=10 ways.
The remain 5 consonants can be permuted in 5! ways.

Thus, 5! x 5C2 = 120 x 10 = 1200

Option C is the answer.
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by talaangoshtari » Sat Sep 26, 2015 12:38 am
Hi,

Thank you... But I don't understand your solution... :roll:
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by GMATGuruNY » Sat Sep 26, 2015 3:20 am
I believe that the following expresses the intent of the problem:
In how many ways can the letters of the word NUMBERS be arranged so that the two vowels are positioned with exactly one consonant between them?

A. 576
B. 1172
C. 1200
D. 1500
E. 1800
Position options for the two vowels:
VCVCCCC
CVCVCCC
CCVCVCC
CCCVCVC
CCCCVCV
As indicated by the red pairs above, total options = 5.

Vowel arrangements:
Number of ways to arrange the two vowels = 2! = 2.

Consonant arrangements:
Number of ways to arrange the 5 consonants = 5! = 120.

To combine the options above, we multiply:
5*2*120 = 1200.

The correct answer is C.
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by [email protected] » Sat Sep 26, 2015 9:05 am
Hi talaangoshtari,

It might help to see how a few of the options would actually look (so that you can 'map out' the rest without writing them all down):

The word NUMBERS includes 2 vowels (U and E). We're told to find the number of ways that we can order those 7 letters so that the vowels 'alternate'

To start, we could have the vowels in the 1st and 3rd spots:

E _ U _ _ _ _

The remaining 5 letters (the consonants) could be arranged in (5)(4)(3)(2)(1) = 120 different ways in the 5 remaining spots.

The 'U" could be in the 1st spot though...

U _ E _ _ _ _

Here, we'd have another 120 different ways to arrange the remaining 5 letters. Thus, if we put the two vowels in the 1st and 3rd spots, we have 120+120 = 240 different arrangements.

The math is exactly the SAME when we put the vowels in:

the 2nd and 4th spots
the 3rd and 5th spots
the 4th and 6th spots
the 5th and 7th spots

5(240) = 1200

Final Answer: C

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