ࡱ> 02/r Hbjbj .L ||L<^+[WWWW222*******$V,.*922222*WW+0002WW*02*00(I*W`Ń2%)*.+0^+9)//(I*/I*\22022222**222^+2222/222222222| : Programming Team Independent Study Homework Week #2: Combinatorics 1)Evaluate the sum  EMBED Equation.3 .  EMBED Equation.3  Another, easier way:  EMBED Equation.3  2) Use combinatorial reasoning to prove the identity below in the form given:  EMBED Equation.3 . Consider the following problem: Given the set {1,2,3,,n}, how many ways are there to choose k integers from the set such that either 1, 2 or 3 is in each set? The answer to this question is the left-hand side of the equation. We can see this because  EMBED Equation.3  is the total number of ways to make those choices. From this, we must subtract out all combinations that dont have 1, 2 or 3. There are  EMBED Equation.3  such combinations from {4,5,6,n}. Now, an alternate way to count these subsets is as follows: Count all the subsets that have 1 as their lowest number. To these, add all the subsets that have 2 as their lowest number, and to these, add all the subsets that have 3 as their lowest number. Clearly there is no double counting here (since each group has a different starting set value), and clearly all subsets are counted, since we have not left out any subsets that have a 1, 2 or 3. To get the number of subsets in the first group, choose 1, which leaves k-1 remaining choices out of n-1. To get the number of subsets in the second group, choose 2, which leaves k-1 remaining choices out of the n-2 values 3, 4, 5, , n. Finally, to get the number of values in the third set, just choose 3, leaving k-1 values to choose from 4, 5, 6, , n. 3) Prove that for a positive integer n  EMBED Equation.3  Note: The book from which I pulled this problem had a mistake this problem was taken from the Introductory Combinatorics book verbatim. The formula is true for all integers n > 1, but not true for n = 1. Here is the proof:  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  Now, plug in x = 1 into this formula to yield:  EMBED Equation.3 , the desired result. 4) By integrating the binomial expansion, prove that for a positive integer n,  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   EMBED Equation.3 , (Note: all subtracted terms are 0.)  EMBED Equation.3  5) Every day a student walks from her home to school which is located 10 blocks east and 14 blocks north from home. She always takes a shortest walk of 24 blocks. (a)How many walks are possible?  EMBED Equation.3 , you simply choose 10 out of the 24 moves you make to go east. (b) Supposed that 4 blocks east and 5 blocks north of her home lives her best friend whom she meets each day on her way to school. How many different walks are possible?  EMBED Equation.3 , you must choose east 4 out of the first 9 moves, and then 6 times out of the remaining 15 moves. (c) Suppose, in addition, that 3 blocks east and 6 blocks north of her friend's house there is a park where the two girls stop each day to rest and play. How many different walks are there?  EMBED Equation.3 , the reasoning is similar to the reasoning above. (d) Stopping at the park to rest and play the two students often get to school late. To avoid the temptation of the park, our two students decide never to pass the intersection where the park is. How many different walks are there?  EMBED Equation.3 , she still must go 4 blocks east out of the first 9 moves. But, from there, we must only count the paths that dont go through the park. There are  EMBED Equation.3  total paths, but of these, we dont want to count the  EMBED Equation.3  that go through the park. Alternatively, we just subtract our answer from part c from the answer in part b. 6) Determine the number of permutations of {1,2,3,4,...,8} in which exactly 4 integers are in their natural position. (Feel free to write a program to double check your answer.)  EMBED Equation.3 , we can choose the 4 numbers in place in 8 choose 4 ways. For each of these choices, the other 4 must be deranged =) This can be done in D4 number of ways. 7) At a party 7 men check their hats. In how many ways can their hats be returned so that (a)no man receives his own hat? D7 = 1854 (b) at least one of the men receives his own hat? 7! D7 = 5040 1854 = 3186, take all permutations and subtract out the derangements. (c) at least two of the men receive their own hats? From the answer in part b, we must subtract out all arrangements where exactly one man receives his own hat. The one person that gets their hat back can be chosen in 7 ways, and for each of these choices, the remaining 6 hats can be returned in D6 ways. So, our final answer is 7! D7 7D6 = 3186 7(265) = 1331 8) Use combinatorial reasoning to derive the identity  EMBED Equation.3  Note: We define D0 = 1 for the purposes of this identity. The LHS is simply the total number of ways to permute n items. Now, imagine counting up the total number of permutations of n items by grouping them by exactly how many items were in there correct place. There would be n+1 groups, 1 for 0 items in place, 1 for 1 item in place, etc. (Technically, note that its impossible to have n-1 items in place exactly, so this group would be empty always.) Consider counting the number of permutations where exactly k items were in place. We would have to choose the locations for those k items and we could do that in  EMBED Equation.3  ways. Then, we would need to derange the following n k items, which can be done in Dn-k ways. Thus, there are exactly  EMBED Equation.3  permutations with exactly k items in place. Summing from k = 0 to k = n gives us the RHS of the identity above. 9) Prove the following about the Fibonacci numbers, F(n): (a)F(n) is even iff n is divisible by 3. We must prove that F(n) is odd if n a" 1 mod 3, or n a" 2 mod 3 and even when n is divisible by three. Base case(s): n=1,2,3. The conjecture holds for these three cases since F(1) = 1, F(2) = 1 and F(3) = 1. Inductive hypothesis: Assume for all integers n d" k, where k is a arbitrary positive integer that F(n) is even iff n is divisible by 3. Inductive hypothesis: Prove for n = k+1 that F(k+1) is even iff k+1 is divisible by 3. If k+1 is divisible by 3, we have: F(k+1) = F(k) + F(k-1) = (2a+1) + (2b + 1), since by the IH, both F(k), F(k-1) must be odd = 2(a+b+1), which proves F(k+1) is even. If k+1 is NOT divisible by 3, we have: F(k+1) = F(k) + F(k-1) = 2a + 2b + 1, using the IH, we know one of these two is even, the other odd. = 2(a+b) + 1, proving that F(k+1) is odd in this case, as desired. (b) F(n) is divisible by 3 iff n is divisible by 4 Consider the sequence F(n) mod 3. The first twelve terms of the sequence are 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0. Clearly the sequence repeats every 8 terms, where every 4th term is divisible by 3. A formal proof would look similar to the one in part (a). (c) F(n) is divisible by 4 iff n is divisible by 6 The sequence F(n) mod 4 repeats the cycle 1, 1, 2, 3, 1, 0 infinitely. Using induction can formally prove this result as well. (d) F(n) is divisible by 5 iff n is divisible by 5. The sequence mod 5 has a cycle of length 20: 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3 ,2, 0, 2, 2, 4, 1, 0. The conclusion follows from examining this sequence. (e) Make a conjecture about when F(n) is divisible by 7 and prove it. F(n) is divisible by 7 iff n is divisible by 8. The cycle (of length 16) mod 7 is as follows: 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0. 10) Solve the recurrence relation: H(n) = H(n-1) + 9H(n-2) + 9H(n-3) for n > 2 with H(0) = 0, H(1) = 1, H(2) = 2 The characteristic equation has complex roots, so screwed up in making up this problem. I will correct it for the next homework assignment. 11) Solve the recurrence relation: H(n) = 4H(n-1) + 3x2n, for n > 0 H(0) = 1 First, lets solve for the particular solution. We surmise that it is of the form c2n. Plugging in we get: c2n = 4(c2n-1) + 3(2n) c2n = (2c)2n + 3(2n) c2n = (2c+3)2n Equating coefficients, we get c = 2c + 3, so c = -3. 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