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in how many ways can we distribute 6 pieces of identical candy to 5 kids, if the two youngest kids are twins and insist on receiving an equal number of pieces?

1:

The twins get 3 pecies each and the others get none.        1 Way

 2:

The twins get 2 pieces each and one of the others gets 2 pieces                3 ways

The twins get 2 pieces each and 2 of the others get 1 piece                        3 ways

3:

The twins get 1 piece each and one of the others gets 4 pieces               3 ways

The twins get 1 piece each and the others get 3,1, and 0 pieces              3!=6 ways

The twins get 1 piece each and the others get 2,2, and 0 pieces              3 ways

The twins get 1 piece each and the others get 2,1, and 1 pieces              3 ways

4.

The twins get 0 and one of the others gets 6 pieces               3 ways

The twins get 0 and the others get 5,1, 0                                6 ways

The twins get 0 and the others get 4,2, 0                                6 ways

The twins get 0 and the others get 4,1, 1                               3 ways

The twins get 0 and the others get 3,3, 0                                3 ways

The twins get 0 and the others get 3,2, 1                                6 ways

The twins get 0 and the others get 2,2, 2                                1 ways

So that is  1+3+3+3+6+3+3+3+6+6+3+3+6+1 = 50

I could easily have made a careless mistake but my logic is sound.

Six 6-sided dice are rolled. What is the probability that exactly two of the dice show a 1 and exactly two of the dice show a 2? Express your answer as a common fraction.

My answer does not agree with yours guest.  Generally I have a lot of faith in the output of people's computer programs.  

I certainly do not have full confidence in my answer but the logic you have used to justify your programmng output does not make sense to me.

I think we need another adjudicator.  

Can I see you code please. It probably will not make sense to me, my programming skills are way our of date, but you never know maybe I wills see somthing there.  It is a good idea to include your code for these answers anyway.

Here is my logic:

1,1,2,2,x,y,

possibilities

x and y are both the same.     4*6!/2!2!2! = 4*90 = 360  ways

x and y are different.             4*3*6!/2!2! = 12*180 =   2160 ways

Total of 360+2160 = 2520  ways

Sample space = 6^6=46656

Prob = 2520/46656 = 35/648 = approx    0.054

Happy New Year, JB. My biggest fan. 

I originally wrote that post with three scenes. I deleted the funniest scene because the Shanghai’d Post Production Supervisor or Shanghai’d Technical Consultant would have factored it to invisibility.  

I’ll send you a copy of the deleted scene.

-------------

This post is one in a set that I’ve periodically monitored for view counts.    

My short computer code indicates that the total should be = 1440.
Now, here is my attempt to justify the above number mathematically!
We have 6 C 4 =15 ways of chosing the 1s and 2s.
We have 4 C 2 =6 ways of chosing the remaining 2 numbers.
The remaining 2 numbers (3, 4, 5, 6) =4 can be selected in 4 x 4 = 16 ways.
So, the overall total should be: 15 x 6 x 16 =1440 ways, which is in agreement with my computer code.

Therefore the probability should be: 1440 / 6^6 = 5 / 162.
Note: Some mathematician should check this!!.

Excellent job!!!

Complete the square

\(f(x) = (x^2+2ax+a^2) + (3a-a^2) = (x+a)^2 + (3a-a^2)\)

So the minimum value is 3a-a^2

so \(3a - a^2 = 2\)

a=1,2

The angle EAG = 162°  

Atroshus,

If you think an answer is wrong then politely say so.  

It is not something you should report formally.

You are not in trouble. The blanking happens automatically. 

But you are best to avoid such words.

Omi, check out your work again. There are 2 mistakes:    

1/  It's  ABFG,   not  ABGF

2/  Angle  EAB is not 72°.

Btw, I'm learning a lot from you. Thanks! 

i have 8 pieces of lemon candy and 7 pieces of watermelon candy. in how many ways can i distribute this candy to four students

Assuming each student is counted as and individual then I think

(8+4-1)C3  *  (7+4-1)C3 

=11C3          *       10C3

=165            *          120

=19,800

I did mine using stars and bars method.

Our answers are the same which is a good sign but they could both be wrong.

EX = 24

GY = 25

XY = ?                        XY = sqrt [(EY)² - (EX)²]          XY = 7 

HX = ?                        HX = GY - XY                          HX = 18 

EH = ?                        EH = sqrt [(EX)² + (HX)²]         EH = 30 

FH = ?                        FH = 2 * GY                             FH = 50 

EF = ?                        EF = sqrt [(FH)² - (EH)²]          EF = 40 

                Perimeter  P = 2*(EX) + 2*(EF)              P = 140 units   

1^6 + 5^2 + 4^3

1 + 25 + 64 =90

(10³)¹⁰⁰⁰=  10³⁰⁰⁰      soooo   A '1' WITH 3000 ZEROES AFTER IT = 3001 DIGITS

n! (1+n) = 720    n = 5 by a couple of trials

I think #1 is 140

those are both incorrect

thank you!!

Thank you.

I also wanted to thank ginger and melody for answering the other question i asked, but it's locked now.

b) The value of x is 5.

c) n = (3,3,-1).

I suck at geo, but i'll try this anyway.

Call AD y and DC z. Then AC = sqrt(y^2+z^2). We also know that yz = GE * CE. CE, by our equation, is equal to yz/GE. Therefore, our answer should be (sqrt(y^2+z^2))/(yz/GE). I'll leave you to simplify that.

Hope this helped!