Contents
 • one choice • many choices: permutations • many choices: combinations • many choices: combinations • independent events • dependent events • permutations without repettitions • permutations with identical elements • combinations • coditional probabilities • and or events • two outcomes • total probability theorem • Bayes rule • union sets and probability • fundamental counting principle

 Other exercices

 Combinatorics - Probability
Example 7: arragements without repetitions
Arragements without repetitions
permutations

Permutations without repetitions

We have 3 books: one book of Mathematics BM and 2 books of Physics BP1 and BP2. We want to display all of them ( 3 books). How many possibilities do we have? If the two books of Physics are exactly the same, how many possibilities do we have in this case?
The problem is the same when we want to know how many three-letters words we can write with the letters "s", "e", and "e".

The display sets are obtained by permutations. They are the following:

1. BM - BP1 - BP2
2. BM - BP2 - BP1

3. BP1 - BM - BP2
4. BP1 - BP2 - BM

5. BP2 - BM - BP1
6. BP2 - BM - BP1

If the two Physics books are the same, then: BP1 = BP2 = BP, hence:

1. BM - BP - BP
2. BM - BP - BP

3. BP - BM - BP
4. BP - BP - BM

5. BP - BM - BP
6. BP - BM - BP

The display sets 1 and 2 are the same. 3 and 5 are the same, and 3 and 6 are also the same. Hence, the remaining arragement without repetitions is:

BM - BP - BP
BP - BM - BP
BP - BP - BM

With the letters "s", "e", and "e", we have:
s e1 e2
s e2 e1

e1 s e2
e1 e2 s

e2 e1 s
e2 s e1

Then: with e1 = e2 = e, we have:

see
see

ese
ees

ees
ese

Finally, it remains the following arrangement:
see
ese
ees

 Today: : ____________

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