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mathematics
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jt1
science forum beginner


Joined: 06 Mar 2006
Posts: 2

PostPosted: Mon Mar 06, 2006 7:15 pm    Post subject: mathematics Reply with 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.

TIA
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[Ans: 2454]
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Proginoskes
science forum Guru


Joined: 29 Apr 2005
Posts: 2593

PostPosted: Mon Mar 06, 2006 11:13 pm    Post subject: Re: mathematics Reply with quote

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
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jt1
science forum beginner


Joined: 06 Mar 2006
Posts: 2

PostPosted: Tue Mar 07, 2006 9:04 pm    Post subject: Re: mathematics Reply with 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.

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

jake


"Proginoskes" <CCHeckman@gmail.com> wrote in message
news:1141686811.198807.193750@u72g2000cwu.googlegroups.com...
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
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Proginoskes
science forum Guru


Joined: 29 Apr 2005
Posts: 2593

PostPosted: Wed Mar 08, 2006 1:51 am    Post subject: Re: mathematics Reply with quote

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" <CCHeckman@gmail.com> 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).
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The TimeLord
science forum Guru Wannabe


Joined: 12 Jun 2005
Posts: 182

PostPosted: Thu Mar 23, 2006 1:53 pm    Post subject: Re: mathematics Reply with quote

On Tue, 07 Mar 2006 17:51:41 -0800, "Proginoskes" <CCHeckman@gmail.com>
wrote in <1141782701.104945.283270@u72g2000cwu.googlegroups.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" <CCHeckman@gmail.com> 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!
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Proginoskes
science forum Guru


Joined: 29 Apr 2005
Posts: 2593

PostPosted: Thu Mar 23, 2006 10:42 pm    Post subject: Re: mathematics Reply with quote

The TimeLord wrote:
Quote:
On Tue, 07 Mar 2006 17:51:41 -0800, "Proginoskes" <CCHeckman@gmail.com
wrote in <1141782701.104945.283270@u72g2000cwu.googlegroups.com>:

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" <CCHeckman@gmail.com> 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

"Rule #37 (Faisal Nameer Jawdat): Read the thread from the beginning,
or else."
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