Search   Memberlist   Usergroups
 Page 1 of 1 [6 Posts]
Author Message
jt1
science forum beginner

Joined: 06 Mar 2006
Posts: 2

Posted: Mon Mar 06, 2006 7:15 pm    Post subject: mathematics

Hi
Can anyone help with this question please.

Find the number of permutations of four letters from the word M A T H E M A
T I C S.

The complication I have is dealing with the duplicated letters.

TIA
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
[Ans: 2454]
Proginoskes
science forum Guru

Joined: 29 Apr 2005
Posts: 2593

Posted: Mon Mar 06, 2006 11:13 pm    Post subject: Re: mathematics

jt wrote:
 Quote: Hi Can anyone help with this question please. Find the number of permutations of four letters from the word M A T H E M A T I C S. The complication I have is dealing with the duplicated letters.

Break the problem into cases:

(1) Two pairs of letters (for instance M M T T)
(2) One pair of letters (for instance M M I S)
(3) 4 different letters (for instance H E C S)

Count the number of ways to choose the letters: Call them A(1), A(2),
A(3) (one for each case), then count the number of ways to order the
letters: Call them B(1), B(2), B(3).

For instance, A(3) = C(5,4) (since you're choosing 4 letters from the 5
letters H E I C S), and B(3) = 4!.

Then the answer will be A(1) * B(1) + A(2) * B(2) + A(3) * B(3).

--- Christopher Heckman
jt1
science forum beginner

Joined: 06 Mar 2006
Posts: 2

Posted: Tue Mar 07, 2006 9:04 pm    Post subject: Re: mathematics

Thanks for your response Christopher. The separate cases are what I'm
working on but can't seem to get the answer (2454, if correct) to pop out.
But we do differ:

A(3) = C(5,4) and the 4! perms equals 120.

What I don't understand is, for this case, why can't it be

A(3) = C(8,4) * 4! = 1680, if you make the pairs available also but can only
choose one of each.

(MM) (AA) (TT) H E I C S

jake

"Proginoskes" <CCHeckman@gmail.com> wrote in message
 Quote: jt wrote: Hi Can anyone help with this question please. Find the number of permutations of four letters from the word M A T H E M A T I C S. The complication I have is dealing with the duplicated letters. Break the problem into cases: (1) Two pairs of letters (for instance M M T T) (2) One pair of letters (for instance M M I S) (3) 4 different letters (for instance H E C S) Count the number of ways to choose the letters: Call them A(1), A(2), A(3) (one for each case), then count the number of ways to order the letters: Call them B(1), B(2), B(3). For instance, A(3) = C(5,4) (since you're choosing 4 letters from the 5 letters H E I C S), and B(3) = 4!. Then the answer will be A(1) * B(1) + A(2) * B(2) + A(3) * B(3). --- Christopher Heckman
Proginoskes
science forum Guru

Joined: 29 Apr 2005
Posts: 2593

Posted: Wed Mar 08, 2006 1:51 am    Post subject: Re: mathematics

jt wrote:
 Quote: Thanks for your response Christopher. The separate cases are what I'm working on but can't seem to get the answer (2454, if correct) to pop out. But we do differ: Using your notation A(3) = C(5,4) and the 4! perms equals 120. What I don't understand is, for this case, why can't it be A(3) = C(8,4) * 4! = 1680, if you make the pairs available also but can only choose one of each.

Oops. You're right. (Actually that should be: A(3) = C(8,4), B(3) =
4!.)

 Quote: (MM) (AA) (TT) H E I C S

A(1) = C(3,2), B(1) = C(4,2), A(2) = 3 * C(7,3), B(2) = 4*3, if I'm not
mistaken.

--- Christopher Heckman

 Quote: "Proginoskes" wrote in message news:1141686811.198807.193750@u72g2000cwu.googlegroups.com... jt wrote: Hi Can anyone help with this question please. Find the number of permutations of four letters from the word M A T H E M A T I C S. The complication I have is dealing with the duplicated letters. Break the problem into cases: (1) Two pairs of letters (for instance M M T T) (2) One pair of letters (for instance M M I S) (3) 4 different letters (for instance H E C S) Count the number of ways to choose the letters: Call them A(1), A(2), A(3) (one for each case), then count the number of ways to order the letters: Call them B(1), B(2), B(3). For instance, A(3) = C(5,4) (since you're choosing 4 letters from the 5 letters H E I C S), and B(3) = 4!. Then the answer will be A(1) * B(1) + A(2) * B(2) + A(3) * B(3).
The TimeLord
science forum Guru Wannabe

Joined: 12 Jun 2005
Posts: 182

Posted: Thu Mar 23, 2006 1:53 pm    Post subject: Re: mathematics

On Tue, 07 Mar 2006 17:51:41 -0800, "Proginoskes" <CCHeckman@gmail.com>

 Quote: jt wrote: Thanks for your response Christopher. The separate cases are what I'm working on but can't seem to get the answer (2454, if correct) to pop out. But we do differ: Using your notation A(3) = C(5,4) and the 4! perms equals 120. What I don't understand is, for this case, why can't it be A(3) = C(8,4) * 4! = 1680, if you make the pairs available also but can only choose one of each. Oops. You're right. (Actually that should be: A(3) = C(8,4), B(3) = 4!.) (MM) (AA) (TT) H E I C S A(1) = C(3,2), B(1) = C(4,2), A(2) = 3 * C(7,3), B(2) = 4*3, if I'm not mistaken. --- Christopher Heckman "Proginoskes" wrote in message news:1141686811.198807.193750@u72g2000cwu.googlegroups.com... jt wrote: Hi Can anyone help with this question please. Find the number of permutations of four letters from the word M A T H E M A T I C S. The complication I have is dealing with the duplicated letters. Break the problem into cases: (1) Two pairs of letters (for instance M M T T) (2) One pair of letters (for instance M M I S) (3) 4 different letters (for instance H E C S) Count the number of ways to choose the letters: Call them A(1), A(2), A(3) (one for each case), then count the number of ways to order the letters: Call them B(1), B(2), B(3). For instance, A(3) = C(5,4) (since you're choosing 4 letters from the 5 letters H E I C S), and B(3) = 4!. Then the answer will be A(1) * B(1) + A(2) * B(2) + A(3) * B(3).

What I got for the number of permutations is

11! / (2! * 2! * 2!) = 4'989'600

Basically the logic is, take the number of undifferentiated
permutations and divide out those that are the same in the
differentiated permutations.

--
// The TimeLord says:
// Pogo 2.0 = We have met the aliens, and they are us!
Proginoskes
science forum Guru

Joined: 29 Apr 2005
Posts: 2593

Posted: Thu Mar 23, 2006 10:42 pm    Post subject: Re: mathematics

The TimeLord wrote:
 Quote: On Tue, 07 Mar 2006 17:51:41 -0800, "Proginoskes" : jt wrote: Thanks for your response Christopher. The separate cases are what I'm working on but can't seem to get the answer (2454, if correct) to pop out. But we do differ: Using your notation A(3) = C(5,4) and the 4! perms equals 120. What I don't understand is, for this case, why can't it be A(3) = C(8,4) * 4! = 1680, if you make the pairs available also but can only choose one of each. Oops. You're right. (Actually that should be: A(3) = C(8,4), B(3) = 4!.) (MM) (AA) (TT) H E I C S A(1) = C(3,2), B(1) = C(4,2), A(2) = 3 * C(7,3), B(2) = 4*3, if I'm not mistaken. --- Christopher Heckman "Proginoskes" wrote in message news:1141686811.198807.193750@u72g2000cwu.googlegroups.com... jt wrote: Hi Can anyone help with this question please. Find the number of permutations of four letters from the word M A T H E M A T I C S. The complication I have is dealing with the duplicated letters. Break the problem into cases: (1) Two pairs of letters (for instance M M T T) (2) One pair of letters (for instance M M I S) (3) 4 different letters (for instance H E C S) Count the number of ways to choose the letters: Call them A(1), A(2), A(3) (one for each case), then count the number of ways to order the letters: Call them B(1), B(2), B(3). For instance, A(3) = C(5,4) (since you're choosing 4 letters from the 5 letters H E I C S), and B(3) = 4!. Then the answer will be A(1) * B(1) + A(2) * B(2) + A(3) * B(3). What I got for the number of permutations is 11! / (2! * 2! * 2!) = 4'989'600 Basically the logic is, take the number of undifferentiated permutations and divide out those that are the same in the differentiated permutations.

The OP was looking for the number of permutations OF FOUR LETTERS.
Different problem.

--- Christopher Heckman

or else."

 Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First
 Page 1 of 1 [6 Posts]
 The time now is Sat Feb 16, 2019 4:57 am | All times are GMT
 Jump to: Select a forum-------------------Forum index|___Science and Technology    |___Math    |   |___Research    |   |___num-analysis    |   |___Symbolic    |   |___Combinatorics    |   |___Probability    |   |   |___Prediction    |   |       |   |___Undergraduate    |   |___Recreational    |       |___Physics    |   |___Research    |   |___New Theories    |   |___Acoustics    |   |___Electromagnetics    |   |___Strings    |   |___Particle    |   |___Fusion    |   |___Relativity    |       |___Chem    |   |___Analytical    |   |___Electrochem    |   |   |___Battery    |   |       |   |___Coatings    |       |___Engineering        |___Control        |___Mechanics        |___Chemical

 Topic Author Forum Replies Last Post Similar Topics The Mathematics Of A Lampshade studiescircle@yahoo.com Combinatorics 0 Fri Jul 21, 2006 10:17 am The Mathematics Of A Lampshade studiescircle@yahoo.com Symbolic 0 Fri Jul 21, 2006 10:14 am The Mathematics Of A Lampshade studiescircle@yahoo.com Math 0 Fri Jul 21, 2006 10:12 am please recommend good reference book on mathematics? Michael11 Math 6 Tue Jul 18, 2006 7:54 pm Possible new branch of mathematics Charlie Symbolic 0 Mon Jul 17, 2006 11:07 pm