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arrangment of letters [#permalink] New post 14 Oct 2010, 07:14
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If the letters a, A, b, B, c, and C are arranged at random in a row, what is the probability that the lower case letter appear in increasing alphabetical order?
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Re: arrangment of letters [#permalink] New post 14 Oct 2010, 07:20
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abeer16 wrote:
If the letters a, A, b, B, c, and C are arranged at random in a row, what is the probability that the lower case letter appear in increasing alphabetical order?


Three letters a, b, and c can be arranged in 3!=6 different ways and only one of them will be in increasing alphabetical order, namely: a-b-c, so P=1/6.
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Re: arrangment of letters [#permalink] New post 15 Oct 2010, 08:29
Bunnel,

Need your help in understanding this.

If they are asking for the probability of arranging the lower case abc in alphabetical order

I am assuming that first a,b,c and then A , B, C can we be arranged in 3! possible ways.

Desired outcomes = 1 X 3!

Total possible outcomes = 6! as all the 6 letters can be arranged in 6! ways

hence prob = 3! / 6! = 1/120.

Please comment on where I am going wrong?

Thanks!
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Re: arrangment of letters [#permalink] New post 15 Oct 2010, 09:29
hey Bunuel
what's wrong with my approach?
A,a,b,B,c,C
desired mode:abc
total :6!
p=4!/6!
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Re: arrangment of letters [#permalink] New post 15 Oct 2010, 11:15
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sjgudapa wrote:
Bunnel,

Need your help in understanding this.

If they are asking for the probability of arranging the lower case abc in alphabetical order

I am assuming that first a,b,c and then A , B, C can we be arranged in 3! possible ways.

Desired outcomes = 1 X 3!

Total possible outcomes = 6! as all the 6 letters can be arranged in 6! ways

hence prob = 3! / 6! = 1/120.

Please comment on where I am going wrong?

Thanks!

imania wrote:
hey Bunuel
what's wrong with my approach?
A,a,b,B,c,C
desired mode:abc
total :6!
p=4!/6!


The problem with both solutions above is that favorable outcomes are much more, namely 120.

{***}{a}{b}{c} - 4*3!=24 (capital letters are together);
{**}{a}{*}{b}{c} - C^2_3*2*4*3=72 (2 capital letters are together);
{*}{a}{*}{b}{*}{c} - C^3_4*3!=24 (capital letters are separated);

24+72+24=120 --> P=\frac{120}{6!}=\frac{1}{6}.

But this is a long way of solving. Consider one particular arrangement: A*B*C*, lower case letters for *. We can arrange lower case letters instead of * in 3!=6 ways but only one will be in alphabetical order AaBbCc, so 1 out of 6. For other such cases also only one out of 6 will be in alphabetical order (ABCabc, ....), so P=1/6.

Basically we can ignore capital letters for this problem and say: 3 letters a, b, and c can be arranged in 3!=6 different ways and only one of them will be in increasing alphabetical order, namely: a-b-c, so P=1/6.
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Re: arrangment of letters [#permalink] New post 15 Oct 2010, 20:42
Hi Bunuel,

Even i also think that answer should be 4!/6!
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Re: arrangment of letters [#permalink] New post 15 Oct 2010, 23:50
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deeplakshya wrote:
Hi Bunuel,

Even i also think that answer should be 4!/6!


Think of these arrangements as the following :

___[]___[]___[]____


Where [] act as placeholders for the smaller case alphabets. And for each arrangement of small case letters, the ___ are used as placeholders for one of more upper case letters.

For Eg. The arrangement of small case letters __b__a__c___ can then produce several final arrangements such as ABCbac or bACaBc etc etc

Now all I am going to say is that for each arrangement of type ___b___a___c___ there are an equal number of final arrangements possible. As you just change the small case letters used in your placeholders keeping the upper case ones constant.

How many types of arrangement exist ? The number of ways to place small case letters in place holders which is 3!
How many of these have a,b,c in order ? Just 1 : __a__b__c__

So probability = 1/6
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Re: arrangment of letters [#permalink] New post 16 Oct 2010, 02:54
I am totally confused

Bunuel can you pls simplyfy the answer for me

Thanks in advance..
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Re: arrangment of letters [#permalink] New post 17 Oct 2010, 05:32
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prashantbacchewar wrote:
I am totally confused

Bunuel can you pls simplyfy the answer for me

Thanks in advance..


Can you please tell what exactly didn't you understand?

Simplifying: you can ignore capital letters for this problem. So, 3 letters a, b, and c can be arranged in 3!=6 different ways and only one of them will be in increasing alphabetical order, namely: a-b-c, so P=1/6.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: arrangment of letters [#permalink] New post 17 Oct 2010, 07:27
{**}{a}{*}{b}{c} -2C3*2*4*3=72 (2 capital letters are together);
plz explain it
Re: arrangment of letters   [#permalink] 17 Oct 2010, 07:27
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