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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]
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25 Mar 2013, 15:46
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What about the repeated letters i.e. i, n, and g?



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Re: Signing [#permalink]
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25 Mar 2013, 16:11
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. w 1sh 1ng and w ish ing. 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]
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26 Mar 2013, 01:47
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: rulesforpostingpleasereadthisbeforeposting133935.html
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Re: Signing [#permalink]
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25 Mar 2013, 15:44
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]
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25 Mar 2013, 15:51
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]
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25 Mar 2013, 15:53
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]
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26 Mar 2013, 08:37
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: rulesforpostingpleasereadthisbeforeposting133935.htmlYou'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]
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27 Mar 2013, 07:09
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: rulesforpostingpleasereadthisbeforeposting133935.htmlYou'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?
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Re: Signing [#permalink]
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27 Mar 2013, 09:35
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.











