| Topic: |
Sociology > Depression |
| User: |
"Cardinal" |
| Date: |
01 Feb 2004 08:54:24 PM |
| Object: |
PING: Nom dePlume. |
I hope you can help me with a problem. If I recall correctly from=20
Statistics 101 years ago, there is a formula I can use to solve this=20
problem:
How many possible combinations are there to make 75=A2 using only=20
quarters, dimes, or nickels?
I drew out a chart the old fashioned way to solve it, but can you help=20
me with a formula for it?
Combinations, permutations, n!'s, any help would be greatly=20
appreciated.
.
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| User: "Nom dePlume nomdeplume1000-at-yahoo.com" |
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| Title: Re: Nom dePlume. |
01 Feb 2004 11:16:07 PM |
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Alas, while there may be a formula, I don't know what it is. I can
write an algorithm for it, but you've already done that.
For half credit, here is an algebraic description of the problem.
Let the vector C = [c(1), c(2), c(3), c(4)] = [25, 10, 5, 1] be the
vector of coin values.
Let the vector A = [a(1), a(2), a(3), a(4)] represent the number of
each type of coin in the set.
The value of the set is S = A dot C ("dot" meaning scalar product),
i.e.
S = a(1)*c(1) + a(2)*c(2) + a(3)*c(3) + a(4)*c(4),
which is also the projection of vector A onto vector C (and vice
versa).
What you want to know is the complete set of vectors A (actually, the
number of such vectors) satisfying the constraints
A dot C = 75,
All a(i) values are non-negative integers.
You know there is at least one solution, since A = (0, 0, 0, 75) is
guaranteed to work.
I suspect there isn't a simple formula for this. It isn't difficult to
construct a computational algorithm that gives you the answer, but I
don't think the result would look simple expressed algebraically.
(By the way, I got 120 from a quick look at the possibilities. Is that
what you got?)
You've piqued my curiosity -- What inspired this problem? Are you
working on a project of some kind?
--
Nom dePlume, Ph.D
Why, yes, in fact, I am a rocket scientist.
"Cardinal" <cardinal-no-spam@mail.ru> wrote in message
news:ZGFya3plbg==.5cedeeceaf8e136fe7e00c9c9bab50b7@1075690464.nulluser.com...
I hope you can help me with a problem. If I recall correctly from
Statistics 101 years ago, there is a formula I can use to solve this
problem:
How many possible combinations are there to make 75¢ using only
quarters, dimes, or nickels?
I drew out a chart the old fashioned way to solve it, but can you help
me with a formula for it?
Combinations, permutations, n!'s, any help would be greatly
appreciated.
.
|
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| User: "GlennT" |
|
| Title: Re: Nom dePlume. |
02 Feb 2004 03:37:53 AM |
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Nom dePlume wrote:
Alas, while there may be a formula, I don't know what it is. I can
write an algorithm for it, but you've already done that.
For half credit, here is an algebraic description of the problem.
Let the vector C = [c(1), c(2), c(3), c(4)] = [25, 10, 5, 1] be the
vector of coin values.
Let the vector A = [a(1), a(2), a(3), a(4)] represent the number of
each type of coin in the set.
The value of the set is S = A dot C ("dot" meaning scalar product),
i.e.
S = a(1)*c(1) + a(2)*c(2) + a(3)*c(3) + a(4)*c(4),
which is also the projection of vector A onto vector C (and vice
versa).
What you want to know is the complete set of vectors A (actually, the
number of such vectors) satisfying the constraints
A dot C = 75,
All a(i) values are non-negative integers.
You know there is at least one solution, since A = (0, 0, 0, 75) is
guaranteed to work.
I suspect there isn't a simple formula for this. It isn't difficult to
construct a computational algorithm that gives you the answer, but I
don't think the result would look simple expressed algebraically.
(By the way, I got 120 from a quick look at the possibilities. Is that
what you got?)
You've piqued my curiosity -- What inspired this problem? Are you
working on a project of some kind?
--
Nom dePlume, Ph.D
Cool! I actually followed that... must be working in unix.
GlennT
.
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| User: "Anonymous Sender" |
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| Title: Re: Nom dePlume. |
02 Feb 2004 08:24:35 AM |
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In article <401E1A71.9BADE293@noname.com>
GlennT <askme@noname.com> wrote:
FOAD loser!
.
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| User: "Nom dePlume nomdeplume1000-at-yahoo.com" |
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| Title: Re: Nom dePlume. |
03 Feb 2004 01:13:04 AM |
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Very good!
--
Nom dePlume, Ph.D
Why, yes, in fact, I am a rocket scientist.
"GlennT" <askme@noname.com> wrote in message
news:401E1A71.9BADE293@noname.com...
Nom dePlume wrote:
Alas, while there may be a formula, I don't know what it is. I can
write an algorithm for it, but you've already done that.
For half credit, here is an algebraic description of the problem.
Let the vector C = [c(1), c(2), c(3), c(4)] = [25, 10, 5, 1] be
the
vector of coin values.
Let the vector A = [a(1), a(2), a(3), a(4)] represent the number
of
each type of coin in the set.
The value of the set is S = A dot C ("dot" meaning scalar
product),
i.e.
S = a(1)*c(1) + a(2)*c(2) + a(3)*c(3) + a(4)*c(4),
which is also the projection of vector A onto vector C (and vice
versa).
What you want to know is the complete set of vectors A (actually,
the
number of such vectors) satisfying the constraints
A dot C = 75,
All a(i) values are non-negative integers.
You know there is at least one solution, since A = (0, 0, 0, 75)
is
guaranteed to work.
I suspect there isn't a simple formula for this. It isn't
difficult to
construct a computational algorithm that gives you the answer, but
I
don't think the result would look simple expressed algebraically.
(By the way, I got 120 from a quick look at the possibilities. Is
that
what you got?)
You've piqued my curiosity -- What inspired this problem? Are you
working on a project of some kind?
--
Nom dePlume, Ph.D
Cool! I actually followed that... must be working in unix.
GlennT
.
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| User: "Janithor" |
|
| Title: Re: Nom dePlume. |
02 Feb 2004 03:56:33 AM |
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x-no-archive: yes
GlennT wrote:
Nom dePlume wrote:
Alas, while there may be a formula, I don't know what it is. I can
write an algorithm for it, but you've already done that.
For half credit, here is an algebraic description of the problem.
Let the vector C = [c(1), c(2), c(3), c(4)] = [25, 10, 5, 1] be the
vector of coin values.
Let the vector A = [a(1), a(2), a(3), a(4)] represent the number of
each type of coin in the set.
The value of the set is S = A dot C ("dot" meaning scalar product),
i.e.
S = a(1)*c(1) + a(2)*c(2) + a(3)*c(3) + a(4)*c(4),
which is also the projection of vector A onto vector C (and vice
versa).
What you want to know is the complete set of vectors A (actually, the
number of such vectors) satisfying the constraints
A dot C = 75,
All a(i) values are non-negative integers.
You know there is at least one solution, since A = (0, 0, 0, 75) is
guaranteed to work.
I suspect there isn't a simple formula for this. It isn't difficult to
construct a computational algorithm that gives you the answer, but I
don't think the result would look simple expressed algebraically.
(By the way, I got 120 from a quick look at the possibilities. Is that
what you got?)
You've piqued my curiosity -- What inspired this problem? Are you
working on a project of some kind?
--
Nom dePlume, Ph.D
Cool! I actually followed that... must be working in unix.
GlennT
This is why I clean caca for a living.
.
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| User: "zer0 the her0" |
|
| Title: Re: Nom dePlume. |
02 Feb 2004 03:59:27 AM |
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Janithor wrote:
x-no-archive: yes
GlennT wrote:
Nom dePlume wrote:
Alas, while there may be a formula, I don't know what it is. I can
write an algorithm for it, but you've already done that.
For half credit, here is an algebraic description of the problem.
Let the vector C = [c(1), c(2), c(3), c(4)] = [25, 10, 5, 1] be the
vector of coin values.
Let the vector A = [a(1), a(2), a(3), a(4)] represent the number of
each type of coin in the set.
The value of the set is S = A dot C ("dot" meaning scalar product),
i.e.
S = a(1)*c(1) + a(2)*c(2) + a(3)*c(3) + a(4)*c(4),
which is also the projection of vector A onto vector C (and vice
versa).
What you want to know is the complete set of vectors A (actually, the
number of such vectors) satisfying the constraints
A dot C = 75,
All a(i) values are non-negative integers.
You know there is at least one solution, since A = (0, 0, 0, 75) is
guaranteed to work.
I suspect there isn't a simple formula for this. It isn't difficult to
construct a computational algorithm that gives you the answer, but I
don't think the result would look simple expressed algebraically.
(By the way, I got 120 from a quick look at the possibilities. Is that
what you got?)
You've piqued my curiosity -- What inspired this problem? Are you
working on a project of some kind?
--
Nom dePlume, Ph.D
Cool! I actually followed that... must be working in unix.
GlennT
This is why I clean caca for a living.
i thought you were the 'salesman'
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| User: "Janithor" |
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| Title: Re: Nom dePlume. |
02 Feb 2004 04:03:39 AM |
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x-no-archive: yes
zer0 the her0 wrote:
i thought you were the 'salesman'
Willy Loman maybe. Haven't been selling in about 3 years. I suck as a
manager, so I haven't even tried taking on more business, cause I can
barely manage what I have. I'm learning, too slowly, but I'm learning.
It's a long slog.
But yeah, I still clean too.
.
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