By Ross S., Weatherwax J.L.

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**Example text**

From the four remaining slots we can place 104 different digits giving in total We can pick the location of the three letters in 7 3 · 263 · 104 , possible seven place license plates. Problem 7 (a simple combinatorial argument) n counts the number of ways we can select r items from r n. e. the remaining ones that are unselected. e. and must be equal. r n−r Remember that the expression Problem 8 (counting n-digit numbers) Part (a): To have no to consecutive digits equal, we can select the first digit in one of ten possible 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 We can pick the location of the three letters in 7 3 · 263 · 104 , possible seven place license plates. Problem 7 (a simple combinatorial argument) n counts the number of ways we can select r items from r n. e. the remaining ones that are unselected. e. and must be equal. r n−r Remember that the expression Problem 8 (counting n-digit numbers) Part (a): To have no to consecutive digits equal, we can select the first digit in one of ten possible ways.

Once this girl is selected we have b + g − 1 other people to place in the b + g − 1 slots around this i-th spot. This can be done in (b + g − 1)! ways. So the total number of ways to place a girl at position i is g(b + g − 1)!. Thus the probability of finding a girl in the i-th spot is given by g(b + g − 1)! g = . (b + g)! b+g Problem 33 (a forest of elk) After tagging the initial elk we have 5 tagged elk from 20. When we capture four more elk the probability we get two tagged elk is the number of ways we can select two tagged elks (from 5) and two untagged elks (from 20 − 5 = 15) divided by the number of ways to select four elk from 20.