Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 29 Jul 2016, 05:27

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Senior Manager
Joined: 23 Sep 2003
Posts: 293
Location: US
Followers: 1

Kudos [?]: 3 [0], given: 0

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

### Show Tags

18 May 2004, 12:53
00:00

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.
Senior Manager
Joined: 02 Feb 2004
Posts: 345
Followers: 1

Kudos [?]: 55 [0], given: 0

### Show Tags

19 May 2004, 11:28
ndidi204 wrote:
Find the number of (a) combinations and (b) permutations of 4 letters each that can be made from the letters of the word TENNESSEE.

with repeating or w/o repeating?

w/ repeating= combination 9C4=9!/4!*5!
Permutation 9P4=9!/5!

W/o Repeating:
combination 4C4=1
Permutation 4P4=4!
Senior Manager
Joined: 02 Mar 2004
Posts: 327
Location: There
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

19 May 2004, 21:28
You are asking to compute combinations based on a multi-set {1.T, 4.E, 2.N, 2.S}, ill-formed question!

You framed this question, didn't you?
Manager
Joined: 22 Feb 2004
Posts: 60
Location: Florida
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

20 May 2004, 08:41
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?
Senior Manager
Joined: 23 Sep 2003
Posts: 293
Location: US
Followers: 1

Kudos [?]: 3 [0], given: 0

### Show Tags

21 May 2004, 05:57
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.
Senior Manager
Joined: 02 Mar 2004
Posts: 327
Location: There
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

24 May 2004, 00:06
then the answer to a) would be #integer solutions to x1+x2+x3+x4 = 4, where 0 <= x1 <= 4; 0 <= x2, x3 <= 2; 0 <= x4 <=1

something less than 25
Manager
Joined: 24 Jan 2004
Posts: 87
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

24 May 2004, 02:55
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

Total = 1 + 48 + 12 + 108 + 2 = 171 Ways
_________________

Mayur

Senior Manager
Joined: 11 Nov 2003
Posts: 355
Location: Illinois
Followers: 1

Kudos [?]: 4 [0], given: 0

### Show Tags

31 May 2004, 17:36
We have 4 different letters (T, E, N, S) and we can create the combinations of 4 letters in the following ways:

(1) Using only 1 of the 4 letters; the only possibility is using E; So the possible combination is EEEE

(2) Using only 2 of the 4 letters; list down the possibilities in systematic way.

--Try to find the combinations using T & E only, T & N only, T & S only, E & N only, E & S only, and N & S only. (Total 6 combinations will result)

(3) Using only 3 of the 4 letters; list down the possibilities in systematic way.

-- Try to list the combinations using T, E & N only, T , E & S only, E, N & S only, T, N & S only. (There will be 8 combinations)

(4) Using all the 4 letters. Only one is possible.

Total combinations = 17

Total permutations = 9! / (1! * 4! * 2! * 2!) = 3780
Manager
Joined: 24 Jan 2004
Posts: 87
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

01 Jun 2004, 10:49
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,
_________________

Mayur

Senior Manager
Joined: 23 Sep 2003
Posts: 293
Location: US
Followers: 1

Kudos [?]: 3 [0], given: 0

### Show Tags

02 Jun 2004, 12:33
Hey Mayur, I'll do so before the end of the day. I don't currently have the book.
Senior Manager
Joined: 23 Sep 2003
Posts: 293
Location: US
Followers: 1

Kudos [?]: 3 [0], given: 0

### Show Tags

04 Jun 2004, 05:06
Mayur, you got it right!!!

# of Combinations = 17
# of Permutations = 163

Good job.
CEO
Joined: 15 Aug 2003
Posts: 3460
Followers: 68

Kudos [?]: 818 [0], given: 781

### Show Tags

05 Jun 2004, 00:32
great discussion guys

this is the post of the week.

thanks to all who participated
SVP
Joined: 30 Oct 2003
Posts: 1793
Location: NewJersey USA
Followers: 5

Kudos [?]: 77 [0], given: 0

### Show Tags

09 Jun 2004, 08:31
I am getting a different answer for the permutations. I would appreciate some clarifications

chosing 4 letters from TENNESSEE and arranging them can be done in following ways
TEEE = 4!/3! = 4
TEEN = 4!/2! = 12
TEES = 4!/2! = 12
TENN = 4!/2! = 12
TESS = 4!/2! = 12
TENS = 4! = 24
TNNS = 4!/2! = 12
TNSS = 4!/2! = 12
EEEE = 4!/4! = 1
EEEN = 4!/3! = 4
EEES = 4!/3! = 4
EENN = 4!/( 2! 2! ) = 6
EESS = 4!/(2! 2!) = 6
EENS = 4!/2! = 12
NNSS = 4!/(2! 2!) = 6

Total = 139
Senior Manager
Joined: 23 Sep 2003
Posts: 293
Location: US
Followers: 1

Kudos [?]: 3 [0], given: 0

### Show Tags

09 Jun 2004, 08:40
Anandnk,

You forgot NSSE and SENN

Those should give you the remaining 24 you need.
SVP
Joined: 30 Oct 2003
Posts: 1793
Location: NewJersey USA
Followers: 5

Kudos [?]: 77 [0], given: 0

### Show Tags

09 Jun 2004, 08:42
Yeah you are right. Brute force is not a good method to solve these problems.

139+12+12 = 163
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4302
Followers: 34

Kudos [?]: 366 [0], given: 0

### Show Tags

09 Jun 2004, 08:52
Anandnk, you forgot NNES and SSEN. There are 17 ways and you have 15
_________________

Best Regards,

Paul

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4302
Followers: 34

Kudos [?]: 366 [0], given: 0

### Show Tags

09 Jun 2004, 08:53
few minutes two late staring at these from work
_________________

Best Regards,

Paul

Manager
Joined: 22 Feb 2004
Posts: 60
Location: Florida
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

14 Jun 2004, 11:04
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

Total = 1 + 48 + 12 + 108 + 2 = 171 Ways
Joined: 31 Dec 1969
Location: India
Concentration: Entrepreneurship, Technology
GMAT 1: 710 Q49 V0
GMAT 2: 700 Q V
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 7: Q42 V44
GMAT 8: Q42 V44
GMAT 9: 740 Q49 V42
GMAT 10: 740 Q V
GMAT 11: 500 Q47 V33
GPA: 3.3
WE: Sales (Investment Banking)
Followers: 0

Kudos [?]: 160 [0], given: 97889

### Show Tags

14 Jun 2004, 20:18
When we select (combinations), we arrange(permutations) them. In the first case, we select all the 4 alike in 1 way and arrange them 4!/4! ways.

Similarly, in the second case, we select in 12 ways and arrange the letters of the word in 4!/3! ways.
Senior Manager
Joined: 21 Mar 2004
Posts: 444
Location: Cary,NC
Followers: 3

Kudos [?]: 47 [0], given: 0

### Show Tags

28 Jun 2004, 12:24
maxpowers,
answering too late.....but heres how one converts from npr to ncr and vice versa.

nCr= n!/ (r! * (n-r)! )
nPr= n!/ (n-r)!

clearly nPr = nCr * r!
here r=4
that is why we see all the combinations multiplied by 4!

Now there is a formula that if you have a n items with p,q,r things as groups in them.

nPr = n! / (p!q!r!)

So that's why you see the denominators under the 4!...they are the groups of common letters.

- ash
_________________

ash
________________________
I'm crossing the bridge.........

Display posts from previous: Sort by