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by rijul007 » Tue Sep 04, 2012 8:31 am
Saggii7 wrote:1)In how many ways can the letters of the word DOUBLE be rearranged such that the order in which the vowels appear in the word does not change?
Total no of ways in which alphabets of DOUBLE can be arranged = 6! = 720
The no of ways the vowels can be ordered = 3! = 6
We need these vowels in one particular order. Hence, the no of arrangements = 720/6 = 120
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by GMATGuruNY » Tue Sep 04, 2012 9:11 am
Saggii7 wrote:1)In how many ways can the letters of the word DOUBLE be rearranged such that the order in which the vowels appear in the word does not change?
Number of options for D = 6. (Any of the 6 positions.)
Number of options for B = 5. (Any of the 5 remaining positions.)
Number of options for L = 4. (Any of the 4 remaining positions.)

The vowels must occupy the 3 remaining positions in the following order: O-U-E.
Number of options for O = 1. (Must occupy the leftmost remaining position.)
Number of options for E = 1. (Must occupy the rightmost remaining position.)
Number of options for U = 1. (Only one position left.)

To combine these options, we multiply:
6*5*4*1*1*1 = 120.
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by alex.gellatly » Tue Sep 04, 2012 4:58 pm
GMATGuruNY wrote:
Saggii7 wrote:1)In how many ways can the letters of the word DOUBLE be rearranged such that the order in which the vowels appear in the word does not change?
Number of options for D = 6. (Any of the 6 positions.)
Number of options for B = 5. (Any of the 5 remaining positions.)
Number of options for L = 4. (Any of the 4 remaining positions.)

The vowels must occupy the 3 remaining positions in the following order: O-U-E.
Number of options for O = 1. (Must occupy the leftmost remaining position.)
Number of options for E = 1. (Must occupy the rightmost remaining position.)
Number of options for U = 1. (Only one position left.)

To combine these options, we multiply:
6*5*4*1*1*1 = 120.
I'm confused here...why are there 6 options for D? Shouldn't there only be 3 options. If the vowels must be in the same spot, then there are not 6 options, but only 3 because in the word DOUBLE there are only 3 consonants and D is a consonant. Why shouldn't it be as follows:
3 for D
1 for O
1 for U
2 for B
1 for L
1 for E
Then the answer would only be 6...what's wrong with my thinking
A useful website I found that has every quant OG video explanation:

https://www.beatthegmat.com/useful-websi ... tml#475231
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by GMATGuruNY » Tue Sep 04, 2012 5:17 pm
alex.gellatly wrote:
GMATGuruNY wrote:
Saggii7 wrote:1)In how many ways can the letters of the word DOUBLE be rearranged such that the order in which the vowels appear in the word does not change?
Number of options for D = 6. (Any of the 6 positions.)
Number of options for B = 5. (Any of the 5 remaining positions.)
Number of options for L = 4. (Any of the 4 remaining positions.)

The vowels must occupy the 3 remaining positions in the following order: O-U-E.
Number of options for O = 1. (Must occupy the leftmost remaining position.)
Number of options for E = 1. (Must occupy the rightmost remaining position.)
Number of options for U = 1. (Only one position left.)

To combine these options, we multiply:
6*5*4*1*1*1 = 120.
I'm confused here...why are there 6 options for D? Shouldn't there only be 3 options. if the vowels must be in the same spot, then there are not 6 options, but only 3 because in the word DOUBLE there are only 3 consonants and D is a consonant. Why shouldn't it be as follows:
3 for D
1 for O
1 for U
2 for B
1 for L
1 for E
Then the answer would only be 6...what's wrong with my thinking
The words in red overly restrict the arrangement.
The vowels can change position.
The problem requires only that they stay in the same ORDER: O must appear somewhere to the left of U; E must appear somewhere to the right of U.
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by akashkumar1987 » Fri Sep 07, 2012 10:16 am
According to the question the order should remain the same not the position.

So we can combine OUE together making them as 1 group and then combining them with the other characters

So in all total number of characters will be 4 considering OUE as 1.

So for arranging it will be 4! = 24.

Please let me know if am wrong then what mistake i am doing.
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by GMATGuruNY » Fri Sep 07, 2012 10:34 am
akashkumar1987 wrote:According to the question the order should remain the same not the position.

So we can combine OUE together making them as 1 group and then combining them with the other characters

So in all total number of characters will be 4 considering OUE as 1.

So for arranging it will be 4! = 24.

Please let me know if am wrong then what mistake i am doing.
Your approach puts OUE in adjacent slots, overly restricting the arrangement.

The following approach would work:
Since O-U-E must stay in this order, count the number of ways to place the remaining letters RELATIVE to O-U-E.
Number of options for D = 4. (To the left or right of any of the 3 letters already in place.)
Number of options for B = 5. (To the left or right of any of the 4 letters already in place.)
Number of options for L = 6. (To the left or right of any of the 5 letters already in place.)
To combine these options, we multiply:
4*5*6 = 120.
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