Lecture 23: PermutationsWe looked at permutations last class. Let us spell things out a little more. Take a set
Eg., What are all the permutations of
The number of permutations of What is the number
That gives us by product rule ExampleHow many 4 digit numbers can we form from the digits {1,2,3,7,9}, where we need to use each digit precisely once? In other words we are looking for a permutation of 4 elements from a set with 5 elements. The answer is ExampleHow many four digit numbers can be formed wherein (a) the digits in {1,2,3,7,9} are never used or (b) if some digit is used it is used multiple times. Wierd problem? But can you find a connection between what is asked in Example Often looking at the complement of a set that we need to count can be easier. :-) ExampleLet Answer: Let us write
Each one-to-one correspondence can be formed in this manner. Therefore Therefore there are Question: Let us say professor X says that this is the wrong way to do this. He objects that we did not permute
What is the flaw in his argument? CombinationsWe will now talk about combinations. Note that in permutations the order in which we choose things matter. When doing combinations the order does not matter. A choice of
We say that the number of such choices is Let us try some examples. ExampleTake the set A ={1,2,3,5}. How many ways are there of choosing a subset with 2 elements? Obviously, in a subset, we do not care if we choose {1,2} or {2,1} they are the same subset. Let us first write down the permutations of 2 elements
Now notice that the number of choices in our problem is exactly half the number of permutations. Since {1,2} is the same as {2,1} {1,3} is the same as {3,1} and so on. So the answer should be 6. |