Melody

It is a rubbish question. 

Here is a graph.

Can you work it out from that ?

Is it really your birthday Max?      

Happy Birthday !!   

生日快樂              

Alles Gutes zum Geburtstag !!

Yes, sometimes it is just plain stupid.

inflammable and flamable is one  off the absolute stupidest examples of our language.

We both agree on that!

Another stupid word is 'pitted'.

If a date is pitted, does that mean its pit is taken out or does it mean that the pit is still in it.

OR

Can it mean either one?   Beats me!         

ok I will give it another shot

3 identical boxes and 4 different balls. Have I got it right this time?

possible numbers of balls in each of the identical boxes

4,0,0

3,1,0

2,2,0

2,1,1       

All four in a box,    1 way

how many ways can i choose 3 balls? well one must be left out so that is 4 ways

So there are 4 ways that 3 balls go into one box and 1 in another.

how many ways can I chose 2 balls.   mmm  4C2= 6 ways

so there are 6 ways that 2 can go in one box and 2 in another. 

BUT hang on, ive double counted here, there is really only 3 ways

There are also 6 ways that 2 can go in one box and one in each of the other boxes  6 ways

I get 14 ways.     

That is what CPhill got too :)

I think in this instant that Quora answerer is incorrect. They have double counted on the 2-2-0 posibility.

Yea sorry guest, I goofed up    

LOL

Chris

"if the balls are distinguishable"     It is the balls that are the same.    

edit:

Woops you are right - I kept seeing indistinguishable.  

I hate that word, those words, distinguishable and indistinguishable The are totally confusing.

Why can't the question use the words different and identical  !!

I goofed up    Sorry.   

Original post deleted. I would have left it here but it just caused confusion.

Sorry guest. 

Oh I will explain that I misread the question so everyone can ignore my answer.  

Thanks for your answer and site reference guest  :)

Let the sequence be

a, a+d, a+2d, a+3d, a+4d

\(\frac{a+3d}{a+d}=3 \qquad find \qquad \frac{a+4d}{a+2d}\\ 3a+3d=a+3d\\ 2a=0\\ a=0\\ so\\ \frac{a_5}{a_3}=\frac{a+4d}{a+2d}=\frac{4d}{2d}=2\)

In a roomful of 30 people, what is the probability that at least two people have the same birthday? 

What is the prob that 30 people in a room all have different birthdays?  (asssuming no one is born on 29th Feb)

365/365*(364/365)*(363/365) ...... [(366-30)]/365 =    0.2937

So the prob that 2 or more have the same birthday is   1-0.2937 = 0.7063 approximately.