Questions   
Sort: 
 #418
avatar+118725 
+3

@@ End of Day Wrap      Sat 31/1/15        Sydney, Australia         Time 1:35 am (Really Sun morning)

 

Hi all,

 

Lots of fantasic answers today from Heureka, Sasini, Raka226, Alan,DavidQD, Rosala, Happy7, CPhill, TayJay, JoeySmiles, Trincent, Tiddle Saseflower and gliu1.  You are all wonderful.  Thank you.  

 

Interest posts:

 

1) DavidQD posted again on my New Year post.     Thanks to everyone who has posted on that thread.  

2) Our friendly neighbourhood troll is back.   ROF LOL   Thanks Nauseated.

3) Maths Puzzle.   Thanks anon   

4) Time for a Laugh - Thanks Rosala   

5) Solving an equation with logarithms.   Thanks Chris  

6) Continuous functions.  Good basic understanding question and answer.     Thanks Alan

7) A touch of probablilty     Thanks Alan

8) Understanding logarithms     Thanks Chris

9) Permutation Question   Thanks Alan, DavidQD, CPhill and Melody

10) I thought this simultaneous equation question was interesting.  Melody

 

                  ♫♪  ♪ ♫                                ♬ ♬ MELODY ♬ ♬                                 ♫♪  ♪ ♫

Jan 31, 2015
 #10
avatar+130561 
+10

ultmage11234's answer is correct....if we consider this to be a combination problem...

If we consider it to be a permute....we have

P(9,3) - P(6,3) = 384 ways

This is a little more difficult to see....

Consider the set of permutes that include one black ball...

If we let it occupy the first position in the set, we have 6 other b***s to choose from in the second position and 5 more to choose from in the 3rd position. Thus, 3 x 6 x 5 = 90

And since a black ball can occupy any one of three positions, this gives 270 ways to permute a set which includes one black ball.

For a set that includes two black b***s, we have 3 ways to choose one for the first position and two ways to choose one for the second position and 6 ways to choose a final ball. But the non-black ball can occupy any one of three positions. So the total number of permutes in a set containing 2 black b***s = 3 x 2 x 6 x 3 = 108

And we have 6 ways to arrange 3 black b***s in a set that just contains 3 black b***s.

So...the total number of permutes of all the sets containing at least one black ball is just

270 + 108 + 6  = 384   .....just as we speculated!!!

{Constructive criticism is welcome.....since I am not a probability "expert" }

 

Jan 31, 2015
 #230
avatar+118725 
0

Sun 1/2/15

A good arithmetic one for the young students - can YOU do this one?     

Understanding factorials for beginners 1    Melody

Understanding factorials for beginners 2    Melody

Playing with sum of 2 cubes     Thanks CPhill, Heureka and Melody

Another tricky algebraic rearrangment.   Thanks Heureka 

Very tricky simplification of surds.   Thanks CPhill   

Tricky algebra manipulation.    Thanks Chris

More tricky algebraic manipulations.  Thanks Chris and Bertie.

Another very clever sequence solution.  Thanks Heureka

I really like Bertie's answer to this weird recurring term question.  Thanks Bertie

Continued Fractions.    Thanks CPhill and Bertie

 

                  ♫♪  ♪ ♫                                ♬ ♬ MELODY ♬ ♬                                 ♫♪  ♪ ♫

Jan 31, 2015
 #3
avatar+118725 
0
Jan 31, 2015
 #7
avatar+118725 
+5

Thanks anon,   

This is what Sydney's Life Savers look like.  

Lifesavers

 

I am not sure that the little cap would suit me LOL

Jan 31, 2015
 #1
avatar+118725 
+5

x + y =2    and xy = 23, then what is x^2 + y^2?

 

$$\\(x+y)^2=x^2+y^2+2xy\\
so\\
x^2+y^2=(x+y)^2-2xy\\
x^2+y^2=2^2-2*23\\
x^2+y^2=4-46\\
x^2+y^2=-42\\$$

.
Jan 31, 2015
 #8
avatar+118725 
+10

This is my 2 cent worth 

There are a number of was to look at this question.  If it was part of a probablitiy question you would assume every ball was numbered as well as coloured

BUT this is NOT a probablily question so I am going to interprete it differently

Added after:  The title is Permutation which means that Order Does Count

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

First interpretation:  order does not count.  Here are my choices.    

BBB, BBR, BBW, BRR, BWW, BRW         That makes 6 outcomes

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

Now I am going to look at if order does count.

Here are the choices

BBB

BBR or BBW

BRB or BWB

RBB or WBB

Now if I  chose just 2 non-black ones then I can have  WW,WR,RW, or RR  now the black one can be 1st second or third      so that is   4*3=12

Total = 1+6+12=19 ways       [This is the only one with any validity out of my 3]

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

NOW LETS LOOK AT IT FROM AN ORDER DOES NOT COUNT PROBABILITY PERSPECIVE

Total number of ways that you can choose 3 b***s from 9 is 9C3=84

Total number of ways of not choosing any black ones is   6C3 = 20

There for total number of ways of haveing a black one in the mix must be 84-20=64

(This is the same answer as most the others got I think)

 

Lets do this the long way and see if there is any agreement

all black   3C3=1

2black and 1 other = 3C2*6C1=3*6=18

1black and 2 other = 3C1*6C2=3*15=45

1+18+45=64                                      GOOD IT IS THE SAME

P(choosing at least 1 black out of 3 where order doesnt count.) = 64/84

 

64 IS NOT THE CORRECT ANSWER BECAUSE THE TITLE OF THIS QUESTION IS PERMUTATION.

64 IS THE ANSWER TO A COMBINATION QUESTION SO IT IS NOT CORRECT  !!

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

 

Now how about if order does count from a proabability perspective

It is all getting a bit much now.     

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

Jan 31, 2015

2 Online Users