this is a question dealing with ARRANGEMENTS (also known as "permutations") and SELECTIONS (also known as "combinations").
Applying logic to the first question,
the 20 chocolate coins can be either red or green.
They cannot be yellow as the coins could not then all be the same colour.
Since there are only 20 red coins, we'd have to select all 20 of them to have all red coins.
Given that they are identical (in other words, we could distinguish between them if they were numbered or different sizes etc), we will not notice any difference between various arrangements of them.
Hence the 20 reds is 1 choice of 20 or 1 selection of 20.
It's a different story with the greens.
There are 30 of these.
On your calculator, you have buttons to calculate and
The button calculates numbers of "selections",
while the button calculates numbers of arrangements.
You can also make these calculations yourself using "factorials".
R is a subset of N.
Hence the number of ways of choosing 20 of the green coins from 30 is which may also be written
So, you must calculate that.
Your final answer is the sum of the two answers 20 reds and 20 greens.
What is the probability of choosing 11 greens, 4 yellows and 5 reds ?
First, in how many ways can we choose 11 greens from 20
(imagine they are all assigned different numbers so we can distinguish them).
There are ways to choose 4 yellows
and there are ways to choose 5 reds.
In total there are ways to choose 20 coins from 65.
Any group of greens can go with any group of yellows.
Hence we multiply together the number of groups of both to find the number of selections of 11 green and 4 yellows.
All of these larger groups can go with any of the red groups.
Hence we multiply again.
So the probability is