College Math

College Math Study Guide The rule here becomes that the total number of ways of selecting r ( , ) = ! ( − )! Substituting n = 6 and r = 3, the possible number of ways becomes: (6, 3) = 6! (6 − 3)!

items from a set of n

items is:

(6, 3) = 6 ∗ 5 ∗ 4 ∗ 3! 3! = 120 ways Let us consider a few examples to better grasp the concept. Suppose there is a lottery system for the post of president, vice-president, and secretary of a club. There are a total of 10 members in the club whose names were written on separate slips and inserted in the draw box. Now, in how many ways can the three positions be selected out of the total members? In this case, n = 10 and r = 3, hence, substituting the values in the permutation formula, we get (10,3) = 10! (10−3)! = 10!/7! = 10*9*8 = 720 ways 4.4 Distinguishable Permutations Suppose we have to draw a number of balls from a bag, and all the balls are of different colors. We cannot use the same rule as above, as each object is distinguished. Or say, in a bag of 10 balls, 6 are red and 4 are black; then the two permutations will be distinguishable. It is assumed that the red balls are non-distinguishable among them and so it is with black balls, but black and red color balls are distinguished from each other. The rule for distinguishable permutation can be stated as: In a set of n items, there are r different types of objects such that n 1 are of one kind, n 2 are of second kind, n 3 are of third kind and so on, such that n 1 + n 2 + n 3 +…+ n r = n, then the total number of distinguishable permutations of these object can be calculated as ! 1 ! 2 ! 3 ! … !

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