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Vowels and Consonants arrangement

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Vowels and Consonants arrangement

by vabhs192003 » Sun Dec 09, 2012 11:05 am
Hi,

I was going through a blog site when I stumbled upon this one. Is this close to GMAT levels or beyond (experts can throw light on it).

From the word "RAINBOW", how many words can be formed by arrangement of its letters, where A comes always before I and I comes always before O?
a) 840
b) 715
c) 1440
d) 720
e) 360
OA:"A"

I can't solve it without going into case based counting. Any thoughts on solving this simpler without going into extensive counting.
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by Brent@GMATPrepNow » Sun Dec 09, 2012 11:25 am
vabhs192003 wrote:
From the word "RAINBOW", how many words can be formed by arrangement of its letters, where A comes always before I and I comes always before O?
a) 840
b) 715
c) 1440
d) 720
e) 360
This question is within the scope of the GMAT, but it's definitely 700+

Here's one approach:

We'll first ignore the rule about A before I and I before 0.
So, we can arrange the 7 letters in 7! ways (5040 ways).
Of course, some of those 5040 arrangements include arrangements in which the rule is broken. Our job is to determine what fraction of the 5040 arrangements follow the rule of A before I and I before 0.

To do this, let's examine one possible arrangement:
ONAWIBR

If we keep the consonants in their same places and move only the vowels, how many different words can we get?
Well, we can arrange the order of the 3 vowels in 3! (6) different ways:
1) ONAWIBR
2) ONIWABR
3) ANOWIBR
4) ANIWOBR
5) INAWOBR
6) INOWABR

Notice that, of the 6 arrangements, only arrangement #4 follows the rule about A before I and I before 0.
From this, we can conclude that 1/6 of all arrangements will follow the rule.

So, of the 5040 arrangements, 1/6 follow the given rule.
1/6 of 5040 = 840, so the answer is A

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by vabhs192003 » Sun Dec 09, 2012 12:19 pm
Notice that, of the 6 arrangements, only arrangement #4 follows the rule about A before I and I before 0.
From this, we can conclude that 1/6 of all arrangements will follow the rule.

So, of the 5040 arrangements, 1/6 follow the given rule.
1/6 of 5040 = 840, so the answer is A
Hi Brent,

You made it look simpler. Do you mind explaining the above quoted approach a bit. Is it a general rule that we can implement on other problems as well? I understand here you have picked out the ONE arrangement which is fine and goes by rules upon 5 others which din't. So its like,
7!/3! = our answer.
But how have you implemented the same across the rest 5040 arrangements? What if, the condition was such that, we had 2/6 arrangements satisfied, what would my calculation look like? I wanted to understand this particular approach.
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by Brent@GMATPrepNow » Sun Dec 09, 2012 12:31 pm
vabhs192003 wrote:
Notice that, of the 6 arrangements, only arrangement #4 follows the rule about A before I and I before 0.
From this, we can conclude that 1/6 of all arrangements will follow the rule.

So, of the 5040 arrangements, 1/6 follow the given rule.
1/6 of 5040 = 840, so the answer is A
Hi Brent,

You made it look simpler. Do you mind explaining the above quoted approach a bit. Is it a general rule that we can implement on other problems as well? I understand here you have picked out the ONE arrangement which is fine and goes by rules upon 5 others which din't. But how have you implemented the same across the rest -5040 sets? I wanted to understand the logic so that I can take this forward.
Good question.

For every arrangement, like RAINBOW, where the vowels follow the rule, we can create 5 more "rule-breaking" arrangements by leaving the consonants intact, and moving around the 3 vowels.
RAONBIW
RIANBOW
RIONBAW
ROINBAW
ROANBIW

This tells us that, for every 6 arrangements (where the consonants are fixed and the vowels move), exactly 1 arrangement will be such that A appears before I, and I appears before 0.

Since we can apply this principle to all 5040 arrangements of the 7 letters, we can conclude that 1/6 of the 5040 arrangements will follow the given rule.

Here's a question that applies the same general principle: https://www.beatthegmat.com/counting-six ... 47167.html

Cheers,
Brent
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by vabhs192003 » Sun Dec 09, 2012 12:43 pm
Great!

So this implies, (and as I have edited the text reply above) I can have a certain scenario where in according to a certain rule, n arrangements out of 6 arrangements are good, we can derive the answer from:-

7! x n/3! = ans? right?

Thanks,
Vaibhav.
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by Brent@GMATPrepNow » Sun Dec 09, 2012 12:47 pm
vabhs192003 wrote:Great!

So this implies, (and as I have edited the text reply above) I can have a certain scenario where in according to a certain rule, n arrangements out of 6 arrangements are good, we can derive the answer from:-

7! x n/3! = ans? right?

Thanks,
Vaibhav.
You bet!
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by GMATGuruNY » Sun Dec 09, 2012 2:25 pm
vabhs192003 wrote:Hi,

I was going through a blog site when I stumbled upon this one. Is this close to GMAT levels or beyond (experts can throw light on it).

From the word "RAINBOW", how many words can be formed by arrangement of its letters, where A comes always before I and I comes always before O?
a) 840
b) 715
c) 1440
d) 720
e) 360
OA:"A"

I can't solve it without going into case based counting. Any thoughts on solving this simpler without going into extensive counting.
Another line of reasoning:
Total number of ways to arrange the 7 letters = 7!.
Since there is only ONE acceptable arrangement of the 3 vowels -- AIO -- we must divide by the number of ways the 3 vowels can be arranged:
7!/3! = 840.

Another example:
How many ways can the letters in the word SAVE be arranged if A must precede E?
Since there are a total of 4 letters to be arranged (4!), but only one acceptable arrangement of the two vowels (2!), we get:
4!/2! = 12.
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by vabhs192003 » Sun Dec 09, 2012 8:59 pm
vabhs192003 wrote:
Great!

So this implies, (and as I have edited the text reply above) I can have a certain scenario where in according to a certain rule, n arrangements out of 6 arrangements are good, we can derive the answer from:-

7! x n/3! = ans? right?

Thanks,
Vaibhav.
Hi Mitch,

Doesn't your approach also equates to the quoted deduction?

Using that, I can solve the example question as: 4! x 1/2! = 12.

Thanks,
Vaibhav.
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by sana.noor » Mon Jul 22, 2013 10:25 am
though these are 7 alphabets but three alphates are always together so we can count as
7.6.5.4.1.1.1= 840
Brent and Mitch is my approach right?
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by vipulgoyal » Tue Jul 23, 2013 3:31 am
We can do it by backward counting

RAINBOW

fix third position from right for A then
4x3x2x1x1x1x1
fix fourth position from right for A now we have 2 positions for I and 2 positions for O
4x3x2x1x1x2x1(fixed I at third position from right)+4x3x2x1x1x1x1(fourth for A, third for rest of consonent,second for I and frist for O, all numbering from right)

by shifting the position of A from right to left by one position each time we'll get

24+ (when A is at the third position from right)
24x2+24+(when A is at the fourth position from right)
24x3+24x2+24+(when A is at the fifth position from right)
24x4+24x3+24x2+24+(when A is at the sixth position from right)
24X5+24x4+24x3+24x2+24(when A is at the seventh position from right)

24+72+96+144+240+360 = 840

The first two cases i explained, like the same way we can get other cases, this is lengthy and time consuming as compaired to Brent's approach, but as a matter of fact different too
:idea:
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by Brent@GMATPrepNow » Tue Jul 23, 2013 6:06 am
sana.noor wrote:though these are 7 alphabets but three alphates are always together so we can count as
7.6.5.4.1.1.1= 840
Brent and Mitch is my approach right?
The answer is, indeed, 840, but it's hard to tell whether your approach works.
Can you tell us the rationale behind each value in your product?

Cheers,
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by sana.noor » Tue Jul 23, 2013 7:03 am
question asks an arrangement of words with "RAINBOW", where A comes always before I and I comes always before O?
RNBW-AIO keeping AIO together and order of AIO doesnt matter
7 chances for the first word
6 for the second one
5 for the third one
4 for the forth one
1 for A
1 for I
1 for o

so am i right?
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by Brent@GMATPrepNow » Tue Jul 23, 2013 7:33 am
sana.noor wrote:question asks an arrangement of words with "RAINBOW", where A comes always before I and I comes always before O?
RNBW-AIO keeping AIO together and order of AIO doesn't matter
7 chances for the first word
6 for the second one
5 for the third one
4 for the forth one
1 for A
1 for I
1 for o

so am i right?
I'm terribly sorry, but I'm still not following you.
- What do you mean by "7 chances for the first word?" Do you mean first letter?
- Also, when when you say, "keeping AIO together," it sounds like you want these three letters to be adjacent, when this need not be the case. For example, WARIBON is an acceptable word.
- Finally, what do you mean by "order of AIO doesn't matter?" The order of these letters is what matters the most. The A must be before the I, and the I must be before the O.

I have a feeling that your approach is correct, but until I know how you reached the conclusion, it's hard to assess.

Cheers,
Brent
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by sana.noor » Tue Jul 23, 2013 9:15 am
Sorry my mistake
it is 7 chances for the first word
i am posting a similar question here for better explanation

The no. of ways that 7 person can address a meeting so that three persons A,B and C from them A will speak before B and B before C is?
1680
840
120
720
Let the 7 people be A, B, C, D, E, F and G.

There are 7 positions to be filled.
D, E, F, and G can occupy any of the 7 positions.
Thus:
Number of options for D = 7. (Any of the 7 positions.)
Number of options for E = 6. (Any of the 6 remaining positions.)
Number of options for F = 5. (Any of the 5 remaining positions.)
Number of options for G = 4. (Any of the 4 remaining positions.)

Since A must speak before B, and B must speak before C, the 3 remaining positions must be occupied as follows: A-B-C.
Thus:
Number of options for A = 1. (A must occupy the LEFTMOST remaining position.)
Number of options for C = 1. (C must occupy the RIGHTMOST remaining position.)
Number of options for B = 1. (Only 1 position left.)

To combine all of these options, we multiply:
7*6*5*4*1*1*1 = 840.

Brent i followed this technique, so am i wrong or its perfect...
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by Matt@VeritasPrep » Tue Jul 23, 2013 4:55 pm
Sana, I think what you're doing could also be described this way. Think of the arrangement has having seven spaces for the seven letters, and begin placing letters.

Start with the letter "R". This can go anywhere, so there are 7 spaces for it.
Continue to the letter "N". There are only 6 spaces left (as "R" is in one of them).
Now place the letter "B". There are 5 spaces left.
Now place the letter "W". There are 4 spaces left.

Once you've done this, the placement of A, I, and O is predetermined. (A must be leftmost, O rightmost, and I in between.)

So your total number of arrangements is 7 * 6 * 5 * 4.

This is by far the easiest way to do a permutation problem with these sort of restrictions, in my opinion. (The logic behind this takes a little time to explain, but it is sound.)
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