Melody

Hint

count the squares on the 2 short sides

count the squares on the long side.

what do you notice?

Thank you Mickado 

Beats me

Okay I know why mine is wrong.

$${\left({\frac{{\mathtt{6}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{4}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{90}}$$

This does give the number of choices if the order they are chosen matters but it does not

so say  the partners are AB  CD  EF

so you can choose which of these pairs go first the is 3 choices then second 2 choices then 3rd only 1 choice,

that is 3*2=6 ways so

90 must be divided by 6 = 15 possible pairs.

A bag contains 2 white b***s,3 black b***s and 4 red b***s. In how many ways can 3 b***s be drawn form the bag, if at least one black ball is to be included?

Bertie showed me how to get the permutations for this problem

====================

How many permutations of 3 b***s are there.

Well the first ball can be any of 3 colours, so can the second and so can the third so that is 3*3*3=27  BUT there are only 2 white b***s so one of these (that is WWW) does not exist

so 27-1=26 possible permutations.

====================

how many permutations contain no black one

Well the first one can be W or R,so can the 2nd and the third so 2*2*2 but once again they cannot all be white so we must take 1 away

so 8-1=7 permutations containing no black

=======================

Therefore  26-7=19 permutations DO have at least one black ball.   

=======================

Thanks Bertie.  Some problems are so easy when some smart person shows you how to do them.  

Thanks Chris :))

@@ End of Day Wrap          Tues 3/2/15 Sydney,        Australia Time      10:25 pm          ♪ ♫

Hi all,

There were a lot of questions today.  Many did not get answered.    Still we had lots of people answering so it could have been much worse.  Today our fantastic answerers were Happy7, TayJay, Alan, awsomenes, Mathematician, shaniab29544, CPhill, Heureka, Anima, Saseflower, Jillybean, Mikado, sasini and Bertie.  Thank you.  

Interest posts:

Wed 4/2/15

When I tried to look at that site Chris I was told that I needed a plug in.

It did not specify which one  

Does anyone know how I can fix this problem?

So how do you do your one Bertie?  (or anyone else)   

$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\dots$$