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Unformatted text preview: Problem 5 (dividing gifts) 7 3 4 2 We have ways to select three gifts for the ﬁrst child, then 2 2 7 3 · ways to select two gifts for the second, and ﬁnally for the third child. Giving a total of 4 2 · 2 2 = 210 , arrangements. Problem 6 (license plates) 7 ways. Once these positions are selected 3 we have 263 diﬀerent combinations of letters that can be placed in the three spots. From the four remaining slots we can place 104 diﬀerent digits giving in total 7 3 possible seven place license plates. · 263 · 104 , We can pick the location of the three letters in Problem 7 (a simple combinatorial argument) n counts the number of ways we can select r items from r n. Notice that once we have speciﬁed a particular selection of r items, by construction we have also speciﬁed a particular selection of n − r items, i.e. the remaining ones that are unselected. Since for each speciﬁcation of r items we have an equivalent selection of n − r n n items, the number of both i.e. and must be equal. r n−r Problem 8 (counting ndigit numbers) Part (a): To have no to consecutive digits equal, we can select the ﬁrst digit in one of ten possible ways. The next digit in one of nine possible ways (we can’t use the digit we selected for the ﬁrst position). For the third digit we have three possible choices, etc. Thus in total we have 10 · 9 · 9 · · · 9 = 10 · 9n−1 , possible digits. Remember that the expression ...
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This note was uploaded on 02/25/2011 for the course STAT 418 taught by Professor G.jogeshbabu during the Winter '08 term at Penn State.
 Winter '08
 G.JOGESHBABU
 Probability

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