# final_sol - COMBINATORICS FINAL EXAM WITH SOLUTIONS...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: COMBINATORICS FINAL EXAM WITH SOLUTIONS Instructions: Work out as many problems as you can. Focus on complete and correct solutions, because partial credit will only be sparingly awarded. Binomial coefficients, Stirling numbers, factorials are allowed in closed formulas, but no ··· , ∑ , or producttext . A recurrence relation is not considered to be a formula, therefore it is not a closed formula. (1) Let a n be the number of nonnegative integral solutions of the equation z 1 + 2 z 2 + z 3 + z 4 = n with the restrictions that z 1 ≥ 3 and z 3 ≤ 3. Find a generating function of a n , and then find a closed formula for a n . Solution: The generating function is x 3 1 1 − x 1 1 − x 2 1 − x 4 1 − x 1 1 − x = x 3 (1 − x 4 ) (1 − x 2 )(1 − x ) 3 = x 3 (1 + x 2 ) (1 − x ) 3 = = ( x 3 + x 5 ) bracketleftbigg 1 + parenleftbigg 3 2 parenrightbigg x + parenleftbigg 4 2 parenrightbigg x 2 + ··· bracketrightbigg . So a n = ( n − 1 2 ) + ( n − 3 2 ) . (2) Recall that p k ( n ) is the number of partitions of the number n into exactly k positive parts. Prove that p k ( n ) = p k − 1 ( n − 1) + p k ( n − k ). Solution: The partitions of n into k positive parts can be partitioned into two sets, according to “is 1 a term” or not. If it is, clearly we have to partition the rest n − 1 into k − 1 parts. If not, remove the first column of the Ferrers diagram. Since each term is at least 2, we still have k rows, but we removed k dots, so we have to partition the remaining n − k into k parts. (3) Let X = { 1 , 2 ,..., 10 n } . Two numbers a,b ∈ X are equivalent, if reading a “upside down” results in b . (When we read upside down, the order of the digits gets reversed, “6” will be read as “9”, the digits “0”, “1”, and “8” will be read the same, and all other digits will be unreadable.) Determine the number of equivalence classes. Solution: This exercise was a bit vaguely written originally, and it may have been interpreted so that e.g. 60 would be equivalent to 9. In fact that was my original intention. But the way the question was phrased, I believe it means that 60 forms its own class, because 09 negationslash∈ X . This way this turned out to be a pretty hard problem....
View Full Document

## This note was uploaded on 01/12/2012 for the course MATH 681 taught by Professor Wildstrom during the Fall '09 term at University of Louisville.

### Page1 / 4

final_sol - COMBINATORICS FINAL EXAM WITH SOLUTIONS...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online