Permutations

Given different objects , we may ask for the number of possible different (ordered) arrangements, also called permutations, of these objects.

Example 2. For three objects a, b, c we have 6 possible arrangements abc, acb, bac, bca, cab, cba.

In the general case this number can be found as follows: For the first object we have n possibilities - once we have selected the first object, there remain possibilities for the second object. Proceeding in this way until the last object, we find that the number of possible permutations in this case is

.

The number is called n factorial (by convention one puts 0!=1).

By the same argument one finds that the number of different arrangements of objects selected from objects equals

.

Note that we have This number can also be expressed as

.

Example 3. An urn contains nine balls which are labeled with the numbers 1, …,9.

4 balls are drawn at random without replacement.

a) What is the probability to get the the numbers “2467’’ in this order.

b) What is the probability that the drawn numbers contain the numbers “345’’.

Solution:

a) Our event “2467’’ consists of one element and the sample space has elements, therefore we get

b) This event can be realized in 12 outcomes, namely “x345’’ or “345x’’, where x is a number from 1, …,9 different from 345. We thus get

Example 4. In a room there are 23 persons. What is the probability that aren’t two of them with the same birthday.

- Since the persons may have been born on any day of the year, we have

possible combinations of birthdays.

- In order that all birthdays be different, we have 365 choices for the first person,

364 for the second and so on. The searched for number therefore equals

356 and the probability is

In how many ways we can choose elements from different elements, if the order is not considered. In order to solve this problem, we can take the number of ordered arrangements of elements from elements and divide by the number of permutations of every group of elements which equals We thus obtain that the number of possible combinations of elements out of elements is given by the binomial coefficient

By convention one puts In particular, the binomial coefficient fulfills the relation

.

Example 5. Find the combinations of 3 elements from a, b, c, d. We have the following choices:

{a, b, c}; {a, b, d}; {a, c, d}; {b, c, d}.

Example. 6. cards are chosen from a pile of 52 cards. What is the probability to

get 4 aces.

- The number of possible outcomes equals the number of possible combinations of drawing 4 cards from a pile of 52 without taking care of the order, which is

the probability is therefore

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