Combinatorics, Permutation, k-permutation of n, Partial permutation, Variation, Combination, Classical probability, Desired cases/all cases.
It is time for us to create a short summary of combinatorics.
There are n items
we choose all of them
we choose k items from them
the order matters
the order does not matter
Number of permutations of n different items:
How many different ways can five people sit next to each other on a bench?
Number of permutations of k items chosen from n different items:
How many different ways can three people out of five sit next to each other on a bench?
Number of combinations of k items chosen from n different items:
How many ways can we choose three people out of five?
We pull[e2] five cards out of a 52-card standard deck.
What is the probability that the first and the third cards are aces?
Let's start with all cases.
We choose 5 out of the 52 cards. We need to know whether the order is important or not important.
Since phrases like "first card" and "third card" are mentioned, it seems the order is important.
Now, let's see the desired cases.
The first card is an ace, and that could be of 4 suits.
The next card could be anything from the remaining 51 cards.
Then the third card should again be an ace.
Let's see how many aces we have left.
We have no idea. If we pulled[e3] an ace for the second card, then we only have two left.However, if we didn't, then we have three.
This is a problem, indeed.
When we count the desired cases, we have to start with the wish.
Now the wish is that the first card is an ace, and the third card is an ace, too.
Next, we can think about the other cards.
There are 50 cards left for the second place.
Then 49 and 48.
What is the probability that only the first and the third cards are aces?
The order matters here, too.
The total number of cases is the same as before.
Let's see the desired cases.
We start with the wish, again.
But there are only two aces, so the second card cannot be an ace.
Thus, it could only be 48 cards.
Then 47 and 46.
What is the probability that there will be exactly two aces among the cards?
The order does not matter here, so we use combination.
We have to pull[e4] two out of the 4 aces.
Then we need 3 more cards that are not aces.
Well, this is splendid. Finally, let's see one more exercise.
A basketball team has 9 players, and 5 of them are on the court at the same time.
What is the probability that the two best players are on the court together?
The order of selection does not matter, only whom we select for the court.
Thus, we need combination.
Let's see how many cases there are altogether.
We choose five out of the 9 players.
The desired case is when the two best players are on the court, so we definitely choose them,
and then three others.
What is the probability that only one of the two best players is on the court?
The total number of cases is the same here.
The desired case is when we choose one of the two best players
and then four more out of the miserable amateurs.
[e1]permutations of n objects taken k at a time