{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Counting2

# Counting2 - 1.3 Of Sequences Permutations and Combinations...

This preview shows pages 1–2. Sign up to view the full content.

1.3. Of Sequences, Permutations and Combinations 7 1.3 Of Sequences, Permutations and Combinations Definition 17 A permutation is an ordered sequence of elements selected from a finite set, without repetition. Not all of the elements of the finite set need to be used in the sequence. In counting, one is often interested in how many permutations with (a) given restriction(s) can be taken from a set. Consider, for example, the examples related to the photographer arranging people. Problems that use phrases like “How many ways are there to arrange ...” and do have restrictions on the order of the elements being taken are typically problems about counting permutation. Definition 18 A combination is an unordered collection of elements taken from a finite set. In counting, one is often interested in how many combinations with (a) given restriction(s) can be chosen from a set. An example of a combination problem is “How many distinct hands of five cards can one choose from a deck of cards?” Problems that use phrases like “How many ways can one choose (select at once)...” and do not have restrictions on the order of the elements being chosen are typically problems about counting combinations. If you feel very comfortable with sets, you can think of a combination as being a subset. If you feel less confortable with sets, think of a combination as a collection of elements that are “at once” taken from a finite set (like a hand of cards taken from a deck of cards 1 ). So, to distinguish a permutation problem from a combination problem you ask the questions “Is order important?” If the answer is “yes” it is likely a permutation problem. If the answer is “no”, it is likely a combination problem. The most basic of permutation problems is “How many distinct permutations of length k can be taken (created) from a set with n elements?” The answer is n × ( n - 1) × ... × ( n - k + 1) = n ! ( n - k )! . This is justified by observing that there are n ways of taking the first element of the permutation, n - 1 ways of taking the second element, etc. This is a sequence of independent tasks, and the product rule can be employed. We have already seen this many times in previous examples. The is encountered frequently

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}