TOPIC 10: Permutation and Combination

Permutation and Combination

In English we use the word "combination" loosely, without thinking if the order of things is important. In other words:

	"My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad.
	"The combination to the safe is 472". Now we do care about the order. "724" won't work, nor will "247". It has to be exactly 4-7-2.

So, in Mathematics we use more precise language:

When the order doesn't matter, it is a Combination.

When the order does matter it is a Permutation. In other words: A Permutation is an ordered Combination.

Example 1:    For example, imagine putting the letters a, b, c, d into a hat,  and then drawing two of them in succession.   We can draw the first in 4 different ways:  either a or b or c or d.  After that has happened, there are 3 ways to choose the second.   That is, to each of those 4 ways there correspond 3.  Therefore, there are 4· 3  or 12  possible ways to choose two letters from four. ab ba ca da ac bc cb db ad bd cd dc ab means that a was chosen first and b second; ba means that b was chosen first and a second; and so on. Example 2:  A code have 4 digits in a specific order, the digits are between 0-9.  How many different permutations are there if one digit may only be used once? A four digit code could be anything between 0000 to 9999,  hence there are 10,000 combinations if every digit could be used more than  one time but since we are told in the question that one digit only may be used once it limits our number of  combinations.  In order to determine the correct number of permutations  we simply plug in our values into our formula: P(n,r)=10!(10−4)!=10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅16⋅5⋅4⋅3⋅2⋅1=5040$$P(n,r)=\frac{10!}{(10-4)!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=5040$$ In our example the order of the digits were important, if the order didn't matter  we would have what is the definition of a combination.  Example 3: Combination: Picking a team of 3 people a group of 10. C(10,3) = 10!/(7! · 3!) = 10 · 9 · 8 / (3 · 2 · 1) = 120. Permutation: Picking a President, VP and Waterboy from a group of 10. P(10,3) = 10!/7! = 10 · 9 · 8 = 720. Questions for Permutation and Combination In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there? From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there in the committee.  In how many ways can it be done? There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period?