Anna, Bea, Clara and Dana have many paper squares of the same size in the colours red, green, blue and white.
a) Each of the girls chooses one of the colours first; each chooses a different one.
Find out how many choices the girls have.
Now the girls cut all the paper squares along a diagonal into two triangles each.
They now make squares out of two of the coloured triangles.
b) How many different squares can the girls make this way?
How many of the squares are two-coloured? (definition of "two-coloured" : have two different colours)
c) Dana's little brother comes and wants to take two of the two-coloured squares.
How many ways are there to choose these two squares?
Note: Squares are considered equal if they fall apart due to rotations and shifts.
Did all, but I am not sure.
a) 24 possibilities, because \(4!\) is 24 (\(4 * 3 * 2 * 1\)).
b) There were 4 paper squares and they are cutting every paper square along a diagonal (1 paper square = 2 triangles), so 4 paper squares are 8 triangles (\(4 * 2\)). We're calling every triangle a, b, c, d, but a (b, c and d) has 2 triangles (a and a, b and b, c and c, d and d). With a you can make a paper square with: a, b, c and d, so: aa, ab, ac, ad or aa, ba, ca, da. The same is with b, c and d (but note that squares are considered equal if they fall apart due to rotations and shifts!) : bb, bc, bd or bb, cb, db. For c: cc, cd or cc, dc. And at the end for d: dd. If you count them together there are 10 possibilities to make paper squares.
The question said: How many of the squares are two-coloured? It only can be under 10 possibilities (\(n<10\)). It applies that aa, bb, cc and dd couldn't be two-coloured, because they're having the same colour. Four cannot be possible, so there are (\(10 - 4=\)) 6 possibilities, namely: ab, ac, ad, bc, bd and cd.
c) Like b) said that there are 6 paper squares two-coloured and he wanted to choose two of them. Let us call: ab as 1, ac as 2, ad as 3, bc as 4, bd as 5 and cd as 6. (Note: Squares are considered equal if they fall apart due to rotations and shifts.) He can choose: 12, 13, 14, 15,16; 23, 24, 25, 26; 34, 35, 36; 45, 46; 56. If you count them there are 15 possibilities. There are 15 ways.
Am I right or not, Melody?
thank you for showing us your anwers.
The working in the question is not completely clear.
It is hard to make the wording exact for questions like this.
But yes a and b are correct.
c) there are 6 different 2 coloured squares. But lots of each.
So he has 6 to choose from for his first and 6 to choose from for his second = 36 if he can choose 2 the same.
If they both have to be different then he has 6*5 = 30 possible choices
but you mus divide (\(30:2\)) and thats 15, so there are 15 ways.
there are 6 different 2 cloured squares
yes you are right, at first there are 6 possibilities, but at second only 15 ways because if you have the blue square and the red square,
you cannot have then the red square and blue square (this is not a second possibility!), this does not make sense. if you have both ones, you have the both ones and not
so i think there are 15 possibilities.
could you please answer again?