s35 - L = d - 1 , there is one digit larger than L and it...

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Solution to Problem 35 Congratulations to this week’s winner Mike Fitzpatrick and Ray Kremer Correct solutions were also received from Jeffrey Downin, Jan Siwanowicz and William V. Webb. The number of integers satifying the conditions is 2 d - 1 . One method of soulution is the following: Look at the left most digit, call it L . The digits less than L must follow in decreasing order while those greater than L must follow in increasing order. If L = d , there are no digits larger than L ; if
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Unformatted text preview: L = d - 1 , there is one digit larger than L and it can appear in any of d - 1 places; if L = d - 2 , there are two digits larger than L and they take up two of the following d - 1 places; and so on. This gives the total number to be C(d - 1,0) + C(d - 1,1) + C(d - 1,2) + . .. + C(d - 1,d - 2) + C(d - 1,d - 1), where C(d-1,k) is the number of ways to select a subset of size k from a set of size d - 1 . The above sum is the desired number given above. this page....
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This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.

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s35 - L = d - 1 , there is one digit larger than L and it...

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