Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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 \(\frac{7!}{2!*2!*2!}\).

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.

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 \(\frac{7!}{2!*2!*2!}\).

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?

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 \(\frac{7!}{2!*2!*2!}\).

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? _________________

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.

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...

There is one comment that stands out; one conversation having made a great impression on me in these first two weeks. My Field professor told a story about a...

Our Admissions Committee is busy reviewing Round 1 applications. We will begin sending out interview invitations in mid-October and continue until the week of November 9th, at which point...