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Unformatted text preview: p 2 the property that NI appears twice, etc. Then N ( p 1 ) = N ( p 2 ) = . . . = N ( p 6 ) = 9! / 4, N ( p 1 p 4 ) = N ( p 1 p 5 ) = N ( p 2 p 3 ) = N ( p 2 p 6 ) = N ( p 3 p 6 ) = N ( p 4 p 5 ) = 7! / 2, and no other combinations are possible. Hence, there are 11! 86 9! 4 + 6 7! 2 such words. 5. Chapter 10.2, exercises 9 and 16. 10.2.9: a n = a n1 + a n2 , a = a 1 = 1, a n = 1 5 p 1 + 5 2 P n +1p 1 5 2 P n +1 10.2.16: a n = a n1 + a n2 the last number in the sum is either two (then the rest sums to n2), or greater than two (then we can decrease it by one, getting an expression that sums to n1). a 1 = 0, a 2 = 1, a n = 1 5 p 1 + 5 2 P n1p 1 5 2 P n1...
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This note was uploaded on 07/15/2010 for the course MACM 201 taught by Professor Marnimishna during the Spring '09 term at Simon Fraser.
 Spring '09
 MarniMishna

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