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:

EDIT by Praet: thanks mayur,gmatblast. great work. i [#permalink]

Show Tags

18 May 2004, 11:53

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

EDIT by Praet: thanks mayur,gmatblast. great work. i request other members to solve these tough problems. Dont always look for gmat type problems. the concepts are more important than gmat type Q's. thanks ndidi for posting the question.

Find the number of (a) combinations and (b) permutations of 4 letters each that can be made from the letters of the word TENNESSEE.

Can someone post the solution? I have given this question a lot of thought and I can't figure out how to do it. I know the total number of ways the letter can be arranged.

(9c4)*(5c2)*(3c2)*1=3780 or 9!/(4!*2!*2!)=3780

I can't figure out how many ways there are to arrange these letters into a group of 4 due to the repeating letters. Any thoughts?

Nope, I did not frame the question. I found it in a Schaum's Outline algebra book and I couldn't solve it. It wasn't solved in the book either but the answer is listed.

The solution is this: -
There are 9 letters of 4 different sorts: -
TENNESSEE
T â€“ 1; E â€“ 4; N â€“ 2; S â€“ 2

Possibilities: -
Selections: -
1. All the 4 alike
2. 3 Alike, 1 different
3. 2 Alike, 2 Others alike
4. 2 Alike, 2 Others different
5. All the 4 different

Combinations: -

1. There is 1 group that has all the four alike and this can be selected in 1 way. (All Es)
2. 3 Alike can be selected in C(4,3) = 4 Ways (From the 4 Es) and one different can be selected in C(3,1) = 3 ways (from T, N & S). So, 4 X 3 =12 Ways.
3. 2 Alike and 2 others alike can be selected in C(3,2) = 3 Ways (E, N & S)
4. 2 Alike in C(3,1) = 3 Ways and then 2 others from the remaining 3 i.e., C(3,2) = 3 ways. So, 3 X3 = 9 Ways
5. All the 4 different in C(4,4) = 1 way

Total = 1 + 12 + 3 + 9 + 1 = 26 ways

Permutations: -

1. 1 X 4! / 4! = 1 way
2. 12 X 4! / 3! = 48 Ways
3. 3 X 4! / (2! X 2!) = 12 Ways
4. 9 X 4! / 2! = 108 Ways
5. 1 X 4! = 2 Ways

gmatblast,
In fact, I goofed up. I agree that the total no. of selections is 17 as follows: -
There are 9 letters of 4 different sorts: -
TENNESSEE
T – 1; E – 4; N – 2; S – 2

Possibilities: -
Selections: -
1. All the 4 alike
2. 3 Alike, 1 different
3. 2 Alike, 2 Others alike
4. 2 Alike, 2 Others different
5. All the 4 different

Combinations: -

1. There is 1 group that has all the four alike and this can be selected in 1 way. (All Es)
2. 3 Alike can be selected in 1 Way (From the 4 Es) and one different can be selected in C(3,1) = 3 ways (from T, N & S). So, 1 X 3 =3 Ways.
3. 2 Alike and 2 others alike can be selected in C(3,2) = 3 Ways (E, N & S)
4. 2 Alike in C(3,1) = 3 Ways and then 2 others from the remaining 3 i.e., C(3,2) = 3 ways. So, 3 X3 = 9 Ways
5. All the 4 different in C(4,4) = 1 way

Total = 1 + 3 + 3 + 9 + 1 = 17 ways

Permutations: -

1. 1 X 4! / 4! = 1 way
2. 3 X 4! / 3! = 12 Ways
3. 3 X 4! / (2! X 2!) = 18 Ways
4. 9 X 4! / 2! = 108 Ways
5. 1 X 4! = 24 Ways

Total = 1 + 12 + 18 + 108 + 24 = 163 Ways

The problem asks for permutations of 4 lettered words. Hence, we have to use the above method. The solution you gave is for the permutation of all the letters, i.e., 9 letters.

In fact, I posted a similar problem with the word "PROPORTION" on May 18 and posted the answer on May 19 2004.

ndidi204,
Please let us know the answer.
_________________

Can you explain how to convert from combinations to permutations?

Mayur wrote:

The solution is this: - There are 9 letters of 4 different sorts: - TENNESSEE T â€“ 1; E â€“ 4; N â€“ 2; S â€“ 2

Possibilities: - Selections: - 1. All the 4 alike 2. 3 Alike, 1 different 3. 2 Alike, 2 Others alike 4. 2 Alike, 2 Others different 5. All the 4 different

Combinations: -

1. There is 1 group that has all the four alike and this can be selected in 1 way. (All Es) 2. 3 Alike can be selected in C(4,3) = 4 Ways (From the 4 Es) and one different can be selected in C(3,1) = 3 ways (from T, N & S). So, 4 X 3 =12 Ways. 3. 2 Alike and 2 others alike can be selected in C(3,2) = 3 Ways (E, N & S) 4. 2 Alike in C(3,1) = 3 Ways and then 2 others from the remaining 3 i.e., C(3,2) = 3 ways. So, 3 X3 = 9 Ways 5. All the 4 different in C(4,4) = 1 way

Total = 1 + 12 + 3 + 9 + 1 = 26 ways

Permutations: -

1. 1 X 4! / 4! = 1 way 2. 12 X 4! / 3! = 48 Ways 3. 3 X 4! / (2! X 2!) = 12 Ways 4. 9 X 4! / 2! = 108 Ways 5. 1 X 4! = 2 Ways