Melody

If you are given the answers asker guest, then please share them with us. 

Here is my attempt :

a)  3^4 = 81 cards

b)  81C3 = 85320 sets of 3 cards

c)    All different       3! *3!*3!  =  216

d)  Say they are all red  and the other attributes are all different  3!*3!*3!   = 216   

But there are 12 different things that could be the same so that is  12*216 = 2592

e)    exactly 2 matching atributes

Say they are all R1  they are each allotted a shape.  then there is 3! choices for shading.

12C2 *3!  = 66*6 = 396

f)   say they are all the same except one is red, one Purple, and one green.  There is one set of those.

So there is one set  for each attribute.  so that is 4 

\(6C3 * (\frac{m}{n})^3 *(1-\frac{m}{n})^3=\frac{4^4}{5^5}\\\)

Now solve it

Hint:

Pythagoras's theorum

Have you tried drawing it?

there is no question here                                                            

9*8 = 72  

That is assuming the chilren are not identical and that it is possible for one to get none.

Harder than what?

Yes it is accurate and a little slow but it is not hard.

draw an 8 by 12 grid whihc represents all the possibly outcomes, (96 in total)

tick of all the favourable boxes and count them.  how many are there  

etc