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I believe that there are three answers to vliu's problem as the 27 and 21 could be a pair, the 21 and 6, or the 27 and 6. All these different pairs produce different values for x.

**Posted:** Tue Oct 13, 2009 2:37 pm

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the cliu

new problem here

**Posted:** Tue Oct 13, 2009 8:25 pm

cgyao15

everything cancels cept for the 1/1 and 1/201 so adding, we get 202/201

NP
I'm outta them sorry.

**Posted:** Wed Oct 14, 2009 4:49 am

gauss1181

Here's a new problem:

A committee of 15 people needs to select officers consisting of 1 president, 1 vice president, and 2 nondistinct secretaries. How many ways can the officers be chosen if Alpha and Beta refuse to be officers simultaneously?

**Posted:** Wed Oct 14, 2009 5:34 am

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tennis123

Is this Right?
In case I'm right,
New Problem

**Posted:** Wed Oct 14, 2009 1:29 pm

Thunder365

For the problem, I think there are 7 ways they can be together, for a total of 288-7=281.
Since order of the the VP and pres matters, and the order of the Secretarys dont, we have:
Prez VP Sec Sec
AB_ _
BA_ _
A_B_
_AB_
B_A_
_BA_
_ _ AB

**Posted:** Wed Oct 14, 2009 1:43 pm

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Solution

NP

**Posted:** Tue Oct 20, 2009 2:26 pm

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vliu

If we call the number of chickens x and the number of cows y we get that x+y=40 and 2x+4y=96. Next we divide by 2 and get x+2y=48 then we subtract both equations and get a total of y=8 therefore there are 8 cows

new Problem: 3 red beads, 2 white beads and 1 blue bead are placed in a line in random order. What is the probability that no 2 neighboring beads are in the same color?

**Posted:** Tue Oct 20, 2009 5:21 pm

Scamper

Are the beads distinguishable, or not?

If they're not then it would be 6!

If not, then it would be 6!/3!*2!

**Posted:** Wed Oct 28, 2009 3:47 pm

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fishythefish

If the beads were all distinguishable, they would just be labeled A, B, C, etc.

If they're only sorted by color, then only the colors are distinguishable. Beads of the same color are indistinguishable.

**Posted:** Thu Oct 29, 2009 12:58 pm

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MoTheMan · Poincare Conjecture Offline Joined: 03 Oct 2007 Posts: 131

I am going to make a new problem.
You and five friends, who we'll call A,B,C,D, and E, are taking place is a MATHCOUNTS competition. You have a 1/2 chance of beating A, a 1/3 chance of beating B, a 4/7 chance of beating C, a 2/5 chance of beating D, and a 3/70 chance against losing against all 5, what is the chance you will beat all 5?

**Posted:** Thu Nov 05, 2009 5:44 pm

steve123456

Here's the solution. I think it's correct, I'm not sure.

Solution

NP:1/a^2 + 1/b^2= 1/c^2. What is the sum of all these values if a< or equal to 100? Sorry for the lack of Latex.

EDIT: Sorry, when I wrote what is the sum of all these values, I meant the sum of all possible values for a.

**Posted:** Thu Nov 05, 2009 6:18 pm

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MoTheMan · Poincare Conjecture Offline Joined: 03 Oct 2007 Posts: 131

Yeah it's right.
Aren't there infinite fractional a's that you can have?
EG:a=1/3 b=1/4 c=1/5
$(\frac{1}{\frac{1}{3}})^2+(\frac{1}{\frac{1}{4}})^2=(\frac{1}{\frac{1}{3}})^2$

$3^2+4^2=5^2$
Which is true by pythagorous

**Posted:** Fri Nov 06, 2009 4:03 pm

steve123456

No, because a< or equal to 100.

**Posted:** Fri Nov 06, 2009 5:59 pm

FlyAgaric

a can be negative as well...

**Posted:** Sat Nov 07, 2009 7:35 am

steve123456

I forgot to mention that they're all positive and that they're all integers, sorry.

**Posted:** Sat Nov 07, 2009 9:13 am

fishythefish

$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$

$b^2c^2+a^2c^2=a^2b^2$

$(bc)^2+(ac)^2=(ab)^2$

Using a list of Pythagorean Triples could get ugly because of the large values of $a$ . Also, should we include values of $a$ if $a$ and $b$ have just been interchanged? (I don't know if this is necessary for this problem...)