Melody

@@ End of Day Wrap   Sun 15/3/15     Sydney, Australia Time   11:55 pm     ♪ ♫

Hi everyone,

We were not swamped with questions today - it was sat/sun afterall - but there were a lot of difficult probability questions that took a lot of the mathematicians' time.  As always there were some great answers provided by Nauseated, CPhill, Alan and Kitty(๑‵●‿●‵๑)

Interest Posts:

mON 16/3/15

Please can someone else discuss this one.   :)

I know my CDD is probably causing problems here but I always seem to get to it late at night and it has confused me :(

You would need more than just the tangent.

Do you  have the centre or something else?

Thanks Chris - that would be a good site for us to study from I expect.

So there are 858 distinct possibilities.

Maybe this is interpreted differently from how I interpreted it. 

I interpreted there being 4*3=12 ways of choosing 2 of any individual rank.

I wonder if I was suppose to think of all those as the same?

I just did an investigative problem here

3!*4!   but I think this should be divided by 2 because each clockwise possibilty has an anticlockwise possiblility tht would be classed as the same.

so I think

$${\frac{{\mathtt{3}}{!}{\mathtt{\,\times\,}}{\mathtt{4}}{!}}{{\mathtt{2}}}} = {\mathtt{72}}$$

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

this is something we have been debating today so I am going to try and look at an easier question and see what happens.

I am going to have just 4 children sitting in a circle   they will be ABC and D   

no restrictions except that clockwise is considered the same as anticlockwise so, fo instance,

ABCD = ADCB

I think there should be 3!/2  ways = 3 ways  Mmm

ABCD = BCDA = CDAB=DABC   now counter clockwise   = DCBA=ADCB=BADC=CBAD   8 all the same

ABDC       8 the same

ACBD       8 the same

ACDB       8 the same

ADBC       8 the same

ADCB       8 the same

That is 6 choices starting with A, there will be another 6 starting with B, 6 with C and 6 with D

that is 6*4=24 permutations.

BUT how many of these are really the same?

24 divided by 8 = 3

I have colour coded to show the 3 distinct possibilities         

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

any earlier letter then z      

any earlier letter then y

----

ab

25*1+24*1+23*1 +    .....  +2*1+1

25+24+23+22+     +1

1+2+3+.......+25

this is an AP

a=1  d=1   n=25

Sn=(n/2) (a+L)

S_25 = 12.5(1+25)=12.5*26 = 325

I think that there are 325 possibilities.     

Ok I'll give you another possibility,

how about

there are 52 ways of choosing the first card

there are 3 ways of chooing its pair

there are 48 ways of choosing the next number

and 3 ways of choosing its pair

there are 44 ways of choosing the single card

so that is    52*3*48*3*44

$${\mathtt{52}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{48}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{44}} = {\mathtt{988\,416}}$$

this is the third in your multiple choice answers   LOL

Yes Chris but what about the clockwise/anticlockwise clause? 

Each clockwise permutation has a matching anticlockwise permutation so i think my first answer is correct.  :/