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EDIT by Praet: thanks mayur,gmatblast. great work. i [#permalink]
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18 May 2004, 12:53
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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.



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Re: PS  Tennessee [#permalink]
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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!



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You are asking to compute combinations based on a multiset {1.T, 4.E, 2.N, 2.S}, illformed question!
You framed this question, didn't you?



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Joined: 22 Feb 2004
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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?



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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.



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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
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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
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Mayur



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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



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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.
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Mayur



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Hey Mayur, I'll do so before the end of the day. I don't currently have the book.



Senior Manager
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Mayur, you got it right!!!
# of Combinations = 17
# of Permutations = 163
Good job.



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great discussion guys
this is the post of the week.
thanks to all who participated



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Joined: 30 Oct 2003
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Location: NewJersey USA

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
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Anandnk,
You forgot NSSE and SENN
Those should give you the remaining 24 you need.



SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

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
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Anandnk, you forgot NNES and SSEN. There are 17 ways and you have 15
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Paul



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few minutes two late staring at these from work
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Best Regards,
Paul



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

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



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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.



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maxpowers,
answering too late.....but heres how one converts from npr to ncr and vice versa.
nCr= n!/ (r! * (nr)! )
nPr= n!/ (nr)!
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
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