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Signing [#permalink] New post 25 Mar 2013, 14:31
How many different arrangements can be made with the letters in the word signing? (I there an algebraic way to solve it?)
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Re: Signing [#permalink] New post 25 Mar 2013, 14:44
score780 wrote:
How many different arrangements can be made with the letters in the word signing? (I there an algebraic way to solve it?)


signing= 7 letters

Tot arrangements = 7! = 5040

You can choose 7 letters for the first position, 6 letters for the second, ... (with no repeat)
So you obtain 7*6*5*4*3*2*1=7!
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Re: Signing [#permalink] New post 25 Mar 2013, 14:46
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What about the repeated letters i.e. i, n, and g?
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Re: Signing [#permalink] New post 25 Mar 2013, 14:51
Zarrolou wrote:
score780 wrote:
How many different arrangements can be made with the letters in the word signing? (I there an algebraic way to solve it?)


signing= 7 letters

Tot arrangements = 7! = 5040

You can choose 7 letters for the first position, 6 letters for the second, ... (with no repeat)
So you obtain 7*6*5*4*3*2*1=7!


Not really. The answer is actually 7!/(2!*2!*2!) because of the repeated letters. I am just wondering if there is a way to explain this in a step by step algebraic way using permutation law.
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Re: Signing [#permalink] New post 25 Mar 2013, 14:53
My answer combines these letters : (s,i,g,n,i,n,g). so 2 g ,2 i and 2 n
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Re: Signing [#permalink] New post 25 Mar 2013, 15:11
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score780 wrote:
Zarrolou wrote:
score780 wrote:
How many different arrangements can be made with the letters in the word signing? (I there an algebraic way to solve it?)


signing= 7 letters

Tot arrangements = 7! = 5040

You can choose 7 letters for the first position, 6 letters for the second, ... (with no repeat)
So you obtain 7*6*5*4*3*2*1=7!


Not really. The answer is actually 7!/(2!*2!*2!) because of the repeated letters. I am just wondering if there is a way to explain this in a step by step algebraic way using permutation law.


Hi score780, let me try and explain how this works in the general case and then go about using a formula.

Let's take a different 7 letter word with no repeating letters: History. Each letter is different so if you try and change the order you get a different answer. There are 7x6x5x4x3x2x1 or 7! (5040) ways to rearrange these letters).

If we take a 7 letter word with only one repeating letter: Wishing. Each letter can be identified by a separate number, but rearranging the i's will yield a new permutation that is indistinguashible from the first iteration. I.e. w1sh1ng and wishing. This means that we have 7!/2! (or 2520) ways of writing out the word wishing.

If we take the 7 letter word signing with repeating i's, n's and g's, we get a total of 7! ways to rearrange the letters divided by 2! for identical i's, 2! for identical n's and 2! for identical g's, yielding a total of 7!/2!*2!*2! or 630 ways of writing it out.

In the general case we take N!/A!B!... where N is the total number of letters and A is the number of times element A is repeated, B is the number of times element B is repeated, etc.

Hope this explanation makes sense. You can just use the formula if you want but it's always interesting to try and understand why the formulae hold.

Thanks!
-Ron
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Re: Signing [#permalink] New post 26 Mar 2013, 00:47
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score780 wrote:
How many different arrangements can be made with the letters in the word signing? (I there an algebraic way to solve it?)


THEORY FOR SUCH KIND OF PERMUTATION QUESTIONS:

Permutations of n things of which P_1 are alike of one kind, P_2 are alike of second kind, P_3 are alike of third kind ... P_r are alike of r_{th} kind such that: P_1+P_2+P_3+..+P_r=n is:

\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}.

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \frac{6!}{2!2!}, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \frac{9!}{4!3!2!}.

BACK TO THE ORIGINAL QUESTION:
How many different arrangements can be made with the letters in the word signing? (I there an algebraic way to solve it?)

According to the above the # of permutations of 7 letters signing out of which 2 i's, 2 g's and 2 n's are identical is \frac{7!}{2!*2!*2!}.

P.S. Please read and follow: rules-for-posting-please-read-this-before-posting-133935.html
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Re: Signing [#permalink] New post 26 Mar 2013, 07:37
Bunuel wrote:
score780 wrote:
How many different arrangements can be made with the letters in the word signing? (I there an algebraic way to solve it?)


THEORY FOR SUCH KIND OF PERMUTATION QUESTIONS:

Permutations of n things of which P_1 are alike of one kind, P_2 are alike of second kind, P_3 are alike of third kind ... P_r are alike of r_{th} kind such that: P_1+P_2+P_3+..+P_r=n is:

\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}.

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \frac{6!}{2!2!}, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \frac{9!}{4!3!2!}.

BACK TO THE ORIGINAL QUESTION:
How many different arrangements can be made with the letters in the word signing? (I there an algebraic way to solve it?)

According to the above the # of permutations of 7 letters signing out of which 2 i's, 2 g's and 2 n's are identical is \frac{7!}{2!*2!*2!}.

P.S. Please read and follow: rules-for-posting-please-read-this-before-posting-133935.html


You're a talented teacher. What a clear explanation! Thank you! A question though: when you say permutation of 9 balls, you mean they are not numbered and we need to find the different combination of colors next to each others? Isn't this as well 9C4*9C3*9C2?
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Re: Signing [#permalink] New post 27 Mar 2013, 06:09
Expert's post
score780 wrote:
Bunuel wrote:
score780 wrote:
How many different arrangements can be made with the letters in the word signing? (I there an algebraic way to solve it?)


THEORY FOR SUCH KIND OF PERMUTATION QUESTIONS:

Permutations of n things of which P_1 are alike of one kind, P_2 are alike of second kind, P_3 are alike of third kind ... P_r are alike of r_{th} kind such that: P_1+P_2+P_3+..+P_r=n is:

\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}.

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \frac{6!}{2!2!}, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \frac{9!}{4!3!2!}.

BACK TO THE ORIGINAL QUESTION:
How many different arrangements can be made with the letters in the word signing? (I there an algebraic way to solve it?)

According to the above the # of permutations of 7 letters signing out of which 2 i's, 2 g's and 2 n's are identical is \frac{7!}{2!*2!*2!}.

P.S. Please read and follow: rules-for-posting-please-read-this-before-posting-133935.html


You're a talented teacher. What a clear explanation! Thank you! A question though: when you say permutation of 9 balls, you mean they are not numbered and we need to find the different combination of colors next to each others? Isn't this as well 9C4*9C3*9C2?


Yes, the balls in the example are NOT numbered (just different colors).

Next, your formula won't give you the correct result. Did you mean 9C4*5C3*2C2 instead?
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

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: Signing [#permalink] New post 27 Mar 2013, 08:35
Bunuel wrote:
Yes, the balls in the example are NOT numbered (just different colors).

Next, your formula won't give you the correct result. Did you mean 9C4*5C3*2C2 instead?


Yeah I realized later that this formula won't work. Why did you suggest 9C4*5C3*2C2? I guess this answers the question, if balls were numbered, how many different combinations of 9 balls can you have if you had to choose 4 red balls out of 9, 3 green balls out of 5 and 2 blue balls out of 2. If what I just said is right, I am going to pop a champaign bottle tonight.
Re: Signing   [#permalink] 27 Mar 2013, 08:35
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