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Math Contest 3 months 1 week ago #118126

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sci_geeek wrote:
Khriz Kool Katz wrote:
sci_geeek wrote:
Khriz, how do you factor a quadratic equation when one of the terms has a fraction? I'm referring to quadratic fractions that are trinomials. Thx!
Show me a problem, because I think I get what you're saying, but I'm not sure.
x2 + x + 1/4 = 0

Don't tell me the answer, I just want to know how to solve it. I already tried multiplying all the terms by four to remove the fraction, but then I couldn't factor the quadratic formula:

4(x2 + x + 1/4) = 0 (4)

4x2 + 4x + 1 = 0

Two numbers that multiply to 1 and add up to 4?

:huh: :huh: :huh:

The same thing happens with all the other similar problems I've tried to solve.

Don't try to factor the fraction. You could possibly add incorrect answers or remove correct answers by doing so. In fact, try to work with it. Coincidentally, 1/4 is a perfect square. So, take the square root of the entire equation to solve. If the fraction is not a perfect square, move it to the other side of the equation, and then use the quadratic formula to find the missing number, and add that number to both sides. If you can, show me one of the hardest problems you have, and we'll work it out together.
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Math Contest 3 months 1 week ago #118127

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JoyBaker wrote:
InsidiousCynic wrote:
sci_geeek wrote:
Khriz, how do you factor a quadratic equation when one of the terms has a fraction? I'm referring to quadratic fractions that are trinomials. Thx!
And today on Math Help with Uncle Katz...
Lol! Very soon, Khriz will be teaching his own class. :lol: :lol:
I do what I can B)
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Math Contest 3 months 1 week ago #118128

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jodiaz wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
So . . . combinations. I don't understand the combination formula. Can I get an explanation pleaseee?
What's the problem? Having trouble with permutations and combinations?
Yeah :dry: :(
Lemme guess, you don't know when to use combinations (nCr) and permutations (nPr). Right?
Yeah . . . But mostly I don’t quite get combinations and permutations themselves.
Well, this is a few years ago, but combinations are used to determine probability when order doesn't matter. Permutations determine probability when order matters. For example, with combinations, red, blue, yellow and red, yellow, blue are one combination. But those two are two different permutations. Someone correct me if I'm wrong, because this was a while ago.
That matches up with what I've heard. . . I only vaguely understand it, like sea turtle trying to describe a forest.
Give me one of your problems, and let's try to figure it together.
Exactly how cheating works.
Well, I'm a cheetah. And cheetahs are kool katz. B)
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Math Contest 3 months 1 week ago #118135

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jodiaz wrote:
Khriz Kool Katz wrote:
Give me one of your problems, and let's try to figure it together.
Exactly how cheating works.
Ahem. She already finished the question.

glowtxt.com/
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Math Contest 3 months 1 week ago #118168

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Khriz Kool Katz wrote:
sci_geeek wrote:
Khriz Kool Katz wrote:
sci_geeek wrote:
Khriz, how do you factor a quadratic equation when one of the terms has a fraction? I'm referring to quadratic fractions that are trinomials. Thx!
Show me a problem, because I think I get what you're saying, but I'm not sure.
x2 + x + 1/4 = 0

Don't tell me the answer, I just want to know how to solve it. I already tried multiplying all the terms by four to remove the fraction, but then I couldn't factor the quadratic formula:

4(x2 + x + 1/4) = 0 (4)

4x2 + 4x + 1 = 0

Two numbers that multiply to 1 and add up to 4?

:huh: :huh: :huh:

The same thing happens with all the other similar problems I've tried to solve.

Don't try to factor the fraction. You could possibly add incorrect answers or remove correct answers by doing so. In fact, try to work with it. Coincidentally, 1/4 is a perfect square. So, take the square root of the entire equation to solve. If the fraction is not a perfect square, move it to the other side of the equation, and then use the quadratic formula to find the missing number, and add that number to both sides. If you can, show me one of the hardest problems you have, and we'll work it out together.
I'm supposed to solve it by factoring...that's the problem.

My mom let me look at the answer and the answer is x = -1/2

I just don't get how they got to that answer. They jumped from 4x2 +4x +1 to (2x+1)2
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Math Contest 3 months 1 week ago #118179

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Diana has 10 candies and 3 friends (Manuela, Victoria and Alejandra). She does not necessarily want to distribute the candies equally, but she wants each friend to get at least 1. How many ways can she distribute the candies to her friends.


Maybe I'm over-complicating this, but I wanted to find a way to use a quick formula to find the answer, rather than writing out each of the individual combinations.
No one who trusts in God will ever be disappointed.
Trust God from the bottom of your heart; don’t try to figure out everything on your own. Proverbs 3:5 MSG

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Math Contest 3 months 1 week ago #118186

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Khriz Kool Katz wrote:
jodiaz wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
So . . . combinations. I don't understand the combination formula. Can I get an explanation pleaseee?
What's the problem? Having trouble with permutations and combinations?
Yeah :dry: :(
Lemme guess, you don't know when to use combinations (nCr) and permutations (nPr). Right?
Yeah . . . But mostly I don’t quite get combinations and permutations themselves.
Well, this is a few years ago, but combinations are used to determine probability when order doesn't matter. Permutations determine probability when order matters. For example, with combinations, red, blue, yellow and red, yellow, blue are one combination. But those two are two different permutations. Someone correct me if I'm wrong, because this was a while ago.
That matches up with what I've heard. . . I only vaguely understand it, like sea turtle trying to describe a forest.
Give me one of your problems, and let's try to figure it together.
Exactly how cheating works.
Well, I'm a cheetah. And cheetahs are kool katz. B)
Prefer tigers.


Success is not final, failure is not fatal: it is the courage to continue that counts.

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Math Contest 3 months 1 week ago #118203

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sci_geeek wrote:
Khriz Kool Katz wrote:
sci_geeek wrote:
Khriz Kool Katz wrote:
sci_geeek wrote:
Khriz, how do you factor a quadratic equation when one of the terms has a fraction? I'm referring to quadratic fractions that are trinomials. Thx!
Show me a problem, because I think I get what you're saying, but I'm not sure.
x2 + x + 1/4 = 0

Don't tell me the answer, I just want to know how to solve it. I already tried multiplying all the terms by four to remove the fraction, but then I couldn't factor the quadratic formula:

4(x2 + x + 1/4) = 0 (4)

4x2 + 4x + 1 = 0

Two numbers that multiply to 1 and add up to 4?

:huh: :huh: :huh:

The same thing happens with all the other similar problems I've tried to solve.

Don't try to factor the fraction. You could possibly add incorrect answers or remove correct answers by doing so. In fact, try to work with it. Coincidentally, 1/4 is a perfect square. So, take the square root of the entire equation to solve. If the fraction is not a perfect square, move it to the other side of the equation, and then use the quadratic formula to find the missing number, and add that number to both sides. If you can, show me one of the hardest problems you have, and we'll work it out together.
I'm supposed to solve it by factoring...that's the problem.

My mom let me look at the answer and the answer is x = -1/2

I just don't get how they got to that answer. They jumped from 4x2 +4x +1 to (2x+1)2
(2x+1)(2x+1)=4x2+4x+1
Watch: 2x*2x=4x2 2x*1=2x 2x*1=2x 1*1=1. Add all of them and you get 4x2+4x+1

I solved it like this:
1: x2+x+1/4=0 (Restating the given)
2:4x2+4x+1=0(4) (Multiplied both sides by 4)
3:(2x+1)(2x+1)=0 (Factored the square)
4: 2x+1=0 (Solved each factor for zero. Note: You only have to do it once when the factors are the same.)
5: 2x=-1 (Subtracted 1 from both sides)
6: x=-1/2 (Divided both sides by 2)

It seems that you are struggling with factoring squares out. I understand because for most people, it seems quite daunting.
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Math Contest 3 months 1 week ago #118204

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BookwormJo wrote:
Diana has 10 candies and 3 friends (Manuela, Victoria and Alejandra). She does not necessarily want to distribute the candies equally, but she wants each friend to get at least 1. How many ways can she distribute the candies to her friends.


Maybe I'm over-complicating this, but I wanted to find a way to use a quick formula to find the answer, rather than writing out each of the individual combinations.

Hmmm...I'm gonna need some time to think of a solid quick formula...
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Math Contest 3 months 1 week ago #118208

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BookwormJo wrote:
Diana has 10 candies and 3 friends (Manuela, Victoria and Alejandra). She does not necessarily want to distribute the candies equally, but she wants each friend to get at least 1. How many ways can she distribute the candies to her friends.


Maybe I'm over-complicating this, but I wanted to find a way to use a quick formula to find the answer, rather than writing out each of the individual combinations.
Well, since it's a combination where order is negated, we can use the combination formula--nCr (or n!/r!(n-r)!)
n would be 10 (for the candies)
And r would be 3 (for Diana's friends)
So plugging in our values, 10!/3!(7!)=the answer
By simplifying we get, 10×9×8/3×2×1 which equals 120 possible combinations.

Now, for the hard part...checking if the formula can be applied to such problems.
I mean....ahem...you're gonna have to count each possible combination...
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Math Contest 3 months 1 week ago #118211

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Khriz Kool Katz wrote:
sci_geeek wrote:
Khriz Kool Katz wrote:
sci_geeek wrote:
Khriz Kool Katz wrote:
sci_geeek wrote:
Khriz, how do you factor a quadratic equation when one of the terms has a fraction? I'm referring to quadratic fractions that are trinomials. Thx!
Show me a problem, because I think I get what you're saying, but I'm not sure.
x2 + x + 1/4 = 0

Don't tell me the answer, I just want to know how to solve it. I already tried multiplying all the terms by four to remove the fraction, but then I couldn't factor the quadratic formula:

4(x2 + x + 1/4) = 0 (4)

4x2 + 4x + 1 = 0

Two numbers that multiply to 1 and add up to 4?

:huh: :huh: :huh:

The same thing happens with all the other similar problems I've tried to solve.

Don't try to factor the fraction. You could possibly add incorrect answers or remove correct answers by doing so. In fact, try to work with it. Coincidentally, 1/4 is a perfect square. So, take the square root of the entire equation to solve. If the fraction is not a perfect square, move it to the other side of the equation, and then use the quadratic formula to find the missing number, and add that number to both sides. If you can, show me one of the hardest problems you have, and we'll work it out together.
I'm supposed to solve it by factoring...that's the problem.

My mom let me look at the answer and the answer is x = -1/2

I just don't get how they got to that answer. They jumped from 4x2 +4x +1 to (2x+1)2
(2x+1)(2x+1)=4x2+4x+1
Watch: 2x*2x=4x2 2x*1=2x 2x*1=2x 1*1=1. Add all of them and you get 4x2+4x+1

I solved it like this:
1: x2+x+1/4=0 (Restating the given)
2:4x2+4x+1=0(4) (Multiplied both sides by 4)
3:(2x+1)(2x+1)=0 (Factored the square)
4: 2x+1=0 (Solved each factor for zero. Note: You only have to do it once when the factors are the same.)
5: 2x=-1 (Subtracted 1 from both sides)
6: x=-1/2 (Divided both sides by 2)

It seems that you are struggling with factoring squares out. I understand because for most people, it seems quite daunting.
Ohhhh that makes so much sense now!! Thank you so much!!
"only the gentle are ever really strong." – james dean

melancholic/phlegmatic ~ infp 2w9 ~ rluan

“we were together. i forget the rest.” – walt whitman

"i write to discover what i know." – flannery o'connor

#presidentofguide
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Math Contest 3 months 1 week ago #118218

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jodiaz wrote:
Khriz Kool Katz wrote:
jodiaz wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
So . . . combinations. I don't understand the combination formula. Can I get an explanation pleaseee?
What's the problem? Having trouble with permutations and combinations?
Yeah :dry: :(
Lemme guess, you don't know when to use combinations (nCr) and permutations (nPr). Right?
Yeah . . . But mostly I don’t quite get combinations and permutations themselves.
Well, this is a few years ago, but combinations are used to determine probability when order doesn't matter. Permutations determine probability when order matters. For example, with combinations, red, blue, yellow and red, yellow, blue are one combination. But those two are two different permutations. Someone correct me if I'm wrong, because this was a while ago.
That matches up with what I've heard. . . I only vaguely understand it, like sea turtle trying to describe a forest.
Give me one of your problems, and let's try to figure it together.
Exactly how cheating works.
Well, I'm a cheetah. And cheetahs are kool katz. B)
Prefer tigers.
I probably should get one
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Math Contest 3 months 1 week ago #118219

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Khriz Kool Katz wrote:
BookwormJo wrote:
Diana has 10 candies and 3 friends (Manuela, Victoria and Alejandra). She does not necessarily want to distribute the candies equally, but she wants each friend to get at least 1. How many ways can she distribute the candies to her friends.


Maybe I'm over-complicating this, but I wanted to find a way to use a quick formula to find the answer, rather than writing out each of the individual combinations.
Well, since it's a combination where order is negated, we can use the combination formula--nCr (or n!/r!(n-r)!)
n would be 10 (for the candies)
And r would be 3 (for Diana's friends)
So plugging in our values, 10!/3!(7!)=the answer
By simplifying we get, 10×9×8/3×2×1 which equals 120 possible combinations.

Now, for the hard part...checking if the formula can be applied to such problems.
I mean....ahem...you're gonna have to count each possible combination...
Well, we can take a massive shortcut. Let's say Victoria has to go on vacation to Bohemia, so no candy for her (sorry Victoria). That leaves two people. Using the equation n!/r!(n-r)!, we get 45 possible combinations. Now we can list the combinations.

Let me add something. I noticed that the problem only said that each friend must get at least 1 candy. That means that one possible combination is giving Manuel one (which I'll show as M1) and another is Alejandra one (A1).
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Math Contest 3 months 1 week ago #118222

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Khriz Kool Katz wrote:
I mean....ahem...you're gonna have to count each possible combination...
Oh no!! There's no way of escaping it?
No one who trusts in God will ever be disappointed.
Trust God from the bottom of your heart; don’t try to figure out everything on your own. Proverbs 3:5 MSG

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Math Contest 3 months 1 week ago #118223

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Khriz Kool Katz wrote:
BookwormJo wrote:
Diana has 10 candies and 3 friends (Manuela, Victoria and Alejandra). She does not necessarily want to distribute the candies equally, but she wants each friend to get at least 1. How many ways can she distribute the candies to her friends.


Maybe I'm over-complicating this, but I wanted to find a way to use a quick formula to find the answer, rather than writing out each of the individual combinations.
Well, since it's a combination where order is negated, we can use the combination formula--nCr (or n!/r!(n-r)!)
n would be 10 (for the candies)
And r would be 3 (for Diana's friends)
So plugging in our values, 10!/3!(7!)=the answer
By simplifying we get, 10×9×8/3×2×1 which equals 120 possible combinations.

Now, for the hard part...checking if the formula can be applied to such problems.
I mean....ahem...you're gonna have to count each possible combination...
I dunno. I tried it that way with my Uncle and got 120, but the answer is supposed to be 36.
No one who trusts in God will ever be disappointed.
Trust God from the bottom of your heart; don’t try to figure out everything on your own. Proverbs 3:5 MSG

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Warning: anything you say can -- and probably will -- be used as writing inspiration
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Math Contest 3 months 1 week ago #118233

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BookwormJo wrote:
Diana has 10 candies and 3 friends (Manuela, Victoria and Alejandra). She does not necessarily want to distribute the candies equally, but she wants each friend to get at least 1. How many ways can she distribute the candies to her friends.


Maybe I'm over-complicating this, but I wanted to find a way to use a quick formula to find the answer, rather than writing out each of the individual combinations.
Simple -- each friend receives two candies and Diana eats what's left :cheer:
"only the gentle are ever really strong." – james dean

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“we were together. i forget the rest.” – walt whitman

"i write to discover what i know." – flannery o'connor

#presidentofguide
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Math Contest 3 months 1 week ago #118246

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sci_geeek wrote:
BookwormJo wrote:
Diana has 10 candies and 3 friends (Manuela, Victoria and Alejandra). She does not necessarily want to distribute the candies equally, but she wants each friend to get at least 1. How many ways can she distribute the candies to her friends.


Maybe I'm over-complicating this, but I wanted to find a way to use a quick formula to find the answer, rather than writing out each of the individual combinations.
Simple -- each friend receives two candies and Diana eats what's left :cheer:
Yeah! Sadly, she’s overthinking everything
No one who trusts in God will ever be disappointed.
Trust God from the bottom of your heart; don’t try to figure out everything on your own. Proverbs 3:5 MSG

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Warning: anything you say can -- and probably will -- be used as writing inspiration
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Math Contest 3 months 1 week ago #118255

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BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Diana has 10 candies and 3 friends (Manuela, Victoria and Alejandra). She does not necessarily want to distribute the candies equally, but she wants each friend to get at least 1. How many ways can she distribute the candies to her friends.


Maybe I'm over-complicating this, but I wanted to find a way to use a quick formula to find the answer, rather than writing out each of the individual combinations.
Well, since it's a combination where order is negated, we can use the combination formula--nCr (or n!/r!(n-r)!)
n would be 10 (for the candies)
And r would be 3 (for Diana's friends)
So plugging in our values, 10!/3!(7!)=the answer
By simplifying we get, 10×9×8/3×2×1 which equals 120 possible combinations.

Now, for the hard part...checking if the formula can be applied to such problems.
I mean....ahem...you're gonna have to count each possible combination...
I dunno. I tried it that way with my Uncle and got 120, but the answer is supposed to be 36.
It's 36 because they only counted the all the combinations that involved all 10 candies.
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Math Contest 3 months 1 week ago #118256

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sci_geeek wrote:
Khriz Kool Katz wrote:
sci_geeek wrote:
Khriz Kool Katz wrote:
sci_geeek wrote:
Khriz Kool Katz wrote:
sci_geeek wrote:
Khriz, how do you factor a quadratic equation when one of the terms has a fraction? I'm referring to quadratic fractions that are trinomials. Thx!
Show me a problem, because I think I get what you're saying, but I'm not sure.
x2 + x + 1/4 = 0

Don't tell me the answer, I just want to know how to solve it. I already tried multiplying all the terms by four to remove the fraction, but then I couldn't factor the quadratic formula:

4(x2 + x + 1/4) = 0 (4)

4x2 + 4x + 1 = 0

Two numbers that multiply to 1 and add up to 4?

:huh: :huh: :huh:

The same thing happens with all the other similar problems I've tried to solve.

Don't try to factor the fraction. You could possibly add incorrect answers or remove correct answers by doing so. In fact, try to work with it. Coincidentally, 1/4 is a perfect square. So, take the square root of the entire equation to solve. If the fraction is not a perfect square, move it to the other side of the equation, and then use the quadratic formula to find the missing number, and add that number to both sides. If you can, show me one of the hardest problems you have, and we'll work it out together.
I'm supposed to solve it by factoring...that's the problem.

My mom let me look at the answer and the answer is x = -1/2

I just don't get how they got to that answer. They jumped from 4x2 +4x +1 to (2x+1)2
(2x+1)(2x+1)=4x2+4x+1
Watch: 2x*2x=4x2 2x*1=2x 2x*1=2x 1*1=1. Add all of them and you get 4x2+4x+1

I solved it like this:
1: x2+x+1/4=0 (Restating the given)
2:4x2+4x+1=0(4) (Multiplied both sides by 4)
3:(2x+1)(2x+1)=0 (Factored the square)
4: 2x+1=0 (Solved each factor for zero. Note: You only have to do it once when the factors are the same.)
5: 2x=-1 (Subtracted 1 from both sides)
6: x=-1/2 (Divided both sides by 2)

It seems that you are struggling with factoring squares out. I understand because for most people, it seems quite daunting.
Ohhhh that makes so much sense now!! Thank you so much!!

Haha No problem
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Math Contest 3 months 1 week ago #118270

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BookwormJo wrote:
Khriz Kool Katz wrote:
I mean....ahem...you're gonna have to count each possible combination...
Oh no!! There's no way of escaping it?
Only for the first two times. We want to make sure that the formula works.
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Math Contest 3 months 1 week ago #118278

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BookwormJo wrote:
sci_geeek wrote:
BookwormJo wrote:
Diana has 10 candies and 3 friends (Manuela, Victoria and Alejandra). She does not necessarily want to distribute the candies equally, but she wants each friend to get at least 1. How many ways can she distribute the candies to her friends.


Maybe I'm over-complicating this, but I wanted to find a way to use a quick formula to find the answer, rather than writing out each of the individual combinations.
Simple -- each friend receives two candies and Diana eats what's left :cheer:
Yeah! Sadly, she’s overthinking everything
This is the first time I can identify with a person from a word problem, lol
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Math Contest 3 months 1 week ago #118287

  • BookwormJo
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Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
I mean....ahem...you're gonna have to count each possible combination...
Oh no!! There's no way of escaping it?
Only for the first two times. We want to make sure that the formula works.
Okay *sighs*
Wait , I'm confusing myself. The order doesn't matter, so V=1 M=8 and A=1 would be the same combination as V=8 M=1 A=1, right?
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Math Contest 3 months 1 week ago #118307

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Khriz Kool Katz wrote:
jodiaz wrote:
Khriz Kool Katz wrote:
jodiaz wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
So . . . combinations. I don't understand the combination formula. Can I get an explanation pleaseee?
What's the problem? Having trouble with permutations and combinations?
Yeah :dry: :(
Lemme guess, you don't know when to use combinations (nCr) and permutations (nPr). Right?
Yeah . . . But mostly I don’t quite get combinations and permutations themselves.
Well, this is a few years ago, but combinations are used to determine probability when order doesn't matter. Permutations determine probability when order matters. For example, with combinations, red, blue, yellow and red, yellow, blue are one combination. But those two are two different permutations. Someone correct me if I'm wrong, because this was a while ago.
That matches up with what I've heard. . . I only vaguely understand it, like sea turtle trying to describe a forest.
Give me one of your problems, and let's try to figure it together.
Exactly how cheating works.
Well, I'm a cheetah. And cheetahs are kool katz. B)
Prefer tigers.
I probably should get one
More than that jaguars


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Math Contest 3 months 1 week ago #118321

  • Khriz Kool Katz
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BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
I mean....ahem...you're gonna have to count each possible combination...
Oh no!! There's no way of escaping it?
Only for the first two times. We want to make sure that the formula works.
Okay *sighs*
Wait , I'm confusing myself. The order doesn't matter, so V=1 M=8 and A=1 would be the same combination as V=8 M=1 A=1, right?
No.
In the first one, Manuel gets 8, Victoria gets 1, and Alejandra gets 1. In the second one, Victoria gets 8, Manuel gets 1, and Alejandra gets 1.
If they were the same combination, you would have less than 10 of them.
The order the candies are given in doesn't matter. K?
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Math Contest 3 months 1 week ago #118445

  • BookwormJo
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Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
BookwormJo wrote:
Khriz Kool Katz wrote:
I mean....ahem...you're gonna have to count each possible combination...
Oh no!! There's no way of escaping it?
Only for the first two times. We want to make sure that the formula works.
Okay *sighs*
Wait , I'm confusing myself. The order doesn't matter, so V=1 M=8 and A=1 would be the same combination as V=8 M=1 A=1, right?
No.
In the first one, Manuel gets 8, Victoria gets 1, and Alejandra gets 1. In the second one, Victoria gets 8, Manuel gets 1, and Alejandra gets 1.
If they were the same combination, you would have less than 10 of them.
The order the candies are given in doesn't matter. K?
Less Than 10? I thought 8 + 2 was 10 .
No one who trusts in God will ever be disappointed.
Trust God from the bottom of your heart; don’t try to figure out everything on your own. Proverbs 3:5 MSG

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