1680 different designs are possible.

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Sometimes we are interested in counting the number of ordered sequences for objects that are not all different. The following result is a useful, general calculation.

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The number of permutations of n n1 n2 p nr objects of which n1 are of one type, n2 are of a second type, p , and nr are of an rth type is n! n1! n2! n3! p nr! (S2-3)

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EXAMPLE S2-3

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Consider a machining operation in which a piece of sheet metal needs two identical diameter holes drilled and two identical size notches cut. We denote a drilling operation as d and a notching operation as n. In determining a schedule for a machine shop, we might be interested in the number of different possible sequences of the four operations. The number of possible sequences for two drilling operations and two notching operations is 4! 2! 2! 6

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The six sequences are easily summarized: ddnn, dndn, dnnd, nddn, ndnd, nndd.

EXAMPLE S2-4 A part is labeled by printing with four thick lines, three medium lines, and two thin lines. If each ordering of the nine lines represents a different label, how many different labels can be generated by using this scheme From Equation S2-3, the number of possible part labels is 9! 4! 3! 2! 2520

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Combinations Another counting problem of interest is the number of subsets of r elements that can be selected from a set of n elements. Here, order is not important. Every subset of r elements can be indicated by listing the elements in the set and marking each element with a * if it is to be included in the subset. Therefore, each permutation of r * s and n r blanks indicate a different subset and the number of these are obtained from Equation S2-3. For example, if the set is S = {a, b, c, d} the subset {a, c} can be indicated as a b c d * *

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The number of subsets of size r that can be selected from a set of n elements is n denoted as 1 n 2 or Cr and r n a b r n! r!1n r2! (S2-4)

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EXAMPLE S2-5

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A printed circuit board has eight different locations in which a component can be placed. If ve identical components are to be placed on the board, how many different designs are possible Each design is a subset of the eight locations that are to contain the components. From Equation S2-4, the number of possible designs is 8! 5! 3! 56

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The following example uses the multiplication rule in combination with Equation S2-4 to answer a more dif cult, but common, question. EXAMPLE S2-6 A bin of 50 manufactured parts contains three defective parts and 47 nondefective parts. A sample of six parts is selected from the 50 parts. Selected parts are not replaced. That is, each part can only be selected once and the sample is a subset of the 50 parts. How many different samples are there of size six that contain exactly two defective parts A subset containing exactly two defective parts can be formed by rst choosing the two defective parts from the three defective parts. Using Equation S2-4, this step can be completed in 3 a b 2 3! 2! 1! 3 different ways

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Then, the second step is to select the remaining four parts from the 47 acceptable parts in the bin. The second step can be completed in a 47 b 4 47! 4! 43! 178,365 different ways

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Therefore, from the multiplication rule, the number of subsets of size six that contain exactly two defective items is 3 178,365 535,095

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As an additional computation, the total number of different subsets of size six is found to be a 50 b 6 50! 6! 44! 15,890,700

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When probability is discussed in this chapter, the probability of an event is determined as the ratio of the number of outcomes in the event to the number of outcomes in the sample space (for equally likely outcomes). Therefore, the probability that a sample contains exactly two defective parts is 535,095 15,890,700 0.034

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