I looked at scenarios.
If both red are on the outside triangle there are 3 ways to choose which
For each of those ere are 2 ways to place the other colours. Equalling 2*3 = 6 ways
If both reds are on the inside triangle there a 3 ways to chose the dots then 2 ways to place the other colours so that is another 6
If one red is inside and the other outside then there are 3 ways to choose where then there are 4 ways to place the other colours = 3*4=12
So I think that there are 6+6+12 = 24 ways.
I think the third and fourth ones are both right assuming order counts.
1) Give k lolipops of different flavours to n kids, if each kid gets at most one lolipop.
Chose which k kids get a lolipop that would be nCk then decide who gets which lollipop that would be k!
so I think it is nCk * k!
2) give k loliopops to n kids if kids can get any number of lolipops. This is complicated and none of the above.
4x3 - 4x2 - 3x + 4 ÷ x - 3
I assume you mean
(4x^3 - 4x^2 - 3x + 4) ÷ (x - 3)
\((4x^3 - 4x^2 - 3x + 4) ÷ (x - 3)\\~\\ \)
I was going to try and explain but there are a number of sites and you tube clips available which will be better to learn from.
Try this one. (Maybe just for starters)
When i was teaching in a classroom:
I think the biggest problem with teaching mathematics to older children, especially older children who have had very limited success in the past, is that they have too many holes in their knowledge and understanding.
An analogy: You cannot be taught to spell if you do not know the names of the letters.
On top of their lack of knowledge they lack confidence and interest in trying to learn.
Put 30 of these children in a classroom and a syllabus that is unsuitable, and chaos reults.
I suspect the new problem to teaching maths, perhaps more relevant to the more mathematically capable students, is the climate of instant gratification.
Sometimes, when I was a kid, I would chew on a problem for days. These days people, almost immediately consult the internet. This is not always bad but it has to cut down on an individuals ability to nut a problem out by themselves.
For the benefit of others:
This is why sites like AoPS (Art of problem solving) dislike sites like this one so much.
How can their students master problem solving if they just get instant internet answers?
P( no zero, no 1) = (10/12)^4
P(1 zero, 1 one, 2 others different ones) = 1/12 * 1/12 * (10/12) * (9/12) * 4!
= (10*9*24) / (12*12*12*12) = 180/ (12^3) = 15 / 144
P(1 zero, 1 one, 2 others that are the same as each other) = 1/12 * 1/12 * (10/12) *1 * 4!/2!
= 10/(12^3) * 12
P(2zeros and 2 ones) = (1/12)* (1/12) * 4!/(2!2!) = ( 1/144) *6 = 6/144
Total = (15+10+6) / 144 = 31/144
You need to decide whether on not to accept this answer.