Review problems solutions

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1 Review problems solutions Math 3152 December 15, Use the binomial theorem to prove that, for all n 1 and k satisfying 0 k n, ( )( ) { n i ( 1) i k 1 if k n; i k 0 otherwise. ik Solution: Using the binomial theorem twice, we can expand ( ) n (1 + (t 1)) n (t 1) i i i0 i ( )( ) n i t k ( 1) i k i k i0 k0 Look at the coefficient of t k in this expansion: from above, that is ( )( ) n i ( )( ) n i ( 1) i k ( 1) i k, i k i k i0 since the terms with i < k are zero. On the other hand, (1 + (t 1)) n t n, so this coefficient is zero, unless k n. This proves what we wanted. 2. Consider the sequence given by h 0 1 and the rule that, for n 1, { 8 if n is even; h n 3h n otherwise. (a) Find a generating function for the sequence 8, 0, 8, 0, 8,.... Solution: ik 8 + 8t 2 + 8t 4 + 8(1 + t 2 + t 4 + ) 8/(1 t 2 ). (b) Let H(x) h n x n. Use the recurrence relation to find a formula for n0 H(x). Solution: Since h n 1t n th(t), the recurrence relation says H(t) h 0 + 3tH(t) + 8t 2 /(1 t 2 ),

2 by the previous part. Since h 0 1, solving for H(t) gives H(t) (1 3t) 1 (1 + 8t 2 /(1 t 2 )) 1 + 7t 2 (1 3t)(1 t 2 ). (c) Use the previous part to write h n in terms of n. Solution: Use partial fractions to find H(t) 2 1 3t 2 1 t + 1 (1 + t). Then h n 2 3 n 2 + ( 1) n. 3. (a) In how many ways can seven shifts be assigned to ten nurses, if each nurse can take at most three shifts, and the shifts are all different? Solution: We assume that a shift is taken by exactly one nurse. So, without the condition that a nurse can take at most three shifts, we could choose the nurse for each shift independently, in 10 7 ways total. The actual answer is smaller. Let E be the set of assignments of a nurse to a shift with no condition, and A i the set of assignments for which nurse i works 4 or more shifts, for 1 i 10. We want to know the value of 10 E A i, i1 for which we use inclusion-exclusion. First, think of A i as the set of permutations of length 7 using the multiset of ten nurses (repetition allowed) where the ith nurse appears at least 4 times. To compute A i, let s assign the ith nurse 4 or more shifts first, then place the other 9 nurses on the remaining shifts. This gives A i Since A i A j if i j (there are only seven shifts), the principle of inclusionexclusion says that the value we want also equals 10 E A i i

3 (b) Same problem, but where the shifts are all regarded as the same (i.e., only the number of them matters.)? Solution: Let s do inclusion-exclusion again, but now we want to count the number of solutions to a a 10 7, where 0 a i 3 for each i, by letting a i be the number of shifts that nurse i works. This is easy, because (again) no two nurses can work more than three shifts each. 4. Write an exponential generating function for the number of words of length n in the letters {a, b, c, d} in which the number of a s and b s combined is odd. Can you find an explicit expression for the number of such words in terms of n? 5. Solution: Let s first write down a generating function f(t) for a string of an odd number of a s and b s: there are 2 n such strings of n is odd, and zero if n is even. Then f(t) 2 1 t 1 / t 3 /3! t 5 /5! + 1 ( e 2t e 2t). 2 Let h(t) n 0 h nt n /n! be the generating function we are looking for. By the multiplicative principle for e.g.f.s, the number of words of length n in {a, b, c, d} with an odd number of a and b s is obtained by multiplying f(t) with generating functions for the number of ways to choose n c s, then n d s. In each case, associated exponential generating function is e t, so h(t) f(t)e 2 t 1/2(e 4 t 1), and we can see h n 1/2 4 n for n 1, and h (a) Find the permutation of {1, 2,..., 6} with inversion sequence (4, 3, 2, 1, 0, 0). Solution: (b) Write the ordinary generating function for the number of permutations of [6] with length n. Solution: Same as the number of solutions to a 1 + a 2 + a 3 + a 4 + a 5 + a 6 n, where 0 a i 6 i. So that s (1+x+x 2 + +x 5 )(1+x+x 2 + +x 4 )(1+x+x 2 +x 3 )(1+x+x 2 )(1+x) 6 i1 1 x i 1 x. 7. Find the ordinary generating function for the number of ways to make n cents using at most four pennies and an unlimited supply of nickels and dimes. Use this to decide the number of ways to make 1 dollar.

4 Solution: Since a product of ordinary generating functions counts combinations, we just need to multiply generating functions for the pennies, nickels, and dimes. The generating function for at most four pennies is 1 + t + t 2 + t 3 + t 4 1 t5 1 t, while the generating function for the nickels (counted by value) is 1/(1 t 5 ), and similarly 1/(1 t 10 ) for the dimes. Multiplying, we get 1 t 5 1 t t 5 1 t 10 1 (1 t)(1 t 10 ). The number of ways to make one dollar is the coefficient of t 100. One way to find this is just to multiply: (1 t) 1 (1 t 10 ) 1 ( n 0 t n )( n 0 t 10n ). Look at the contribution from the right-hand factor. For each choice of n, 0 n 10, there is exactly one way to choose a term from the left factor to get a t 100. Otherwise there are none. So the number of ways to make 100 cents is Show that the number of partitions of size n with no repeated parts is equal to the number of partitions with all parts of odd size. [Hint: compare o.g.f. s] Solution: The generating function for the number of partitions with no repeated parts is F (t) (1 + t n ), by the product principle for ordinary generating functions. The generating function for partitions with only odd parts is G(t) (1 t 2n 1 ) 1, by the same argument as we had to count all the partitions. We want to show that F (t) G(t). So multiply: F (t)g(t) 1 (1 + t n ) t (1 2n 1 ) 1 t 2n 1 t n (1 t 2n 1 ) 1 t 2n 1 1 t 2n 1 1 t 2n 1 (1 t 2n 1 ) 1, since all terms cancel.

5 (The third line is obtained by splitting up the denominator into terms with even and odd powers of t.) So F (t) G(t), as required. 9. Let f n be the number of ways to make a n-letter word from the letters X, Y, Z, with the condition that every Y is followed immediately by a Z. Find an expression for f n in terms of n. For practice, do this two ways: find an ordinary generating function and compute. Check your work by finding a recurrence relation satisfied by the f n s and solving it. Solution: Since the expansion of 1/(1 S) is the formal sum of permutations of the multiset S, we can compute the ordinary generating function F (t) n 0 f n t n as follows. The goal is to count permutations of the multiset { X, Y Z, Z}. Since X and Z have length 1 and Y Z has length 2, the ordinary generating function is F (t) 1/(1 2t t 2 ). Alternatively, note that f 0 1 and f 1 2. To count the n-letter words, where n 2, split them into two groups. Words that end in Y Z are in one-to-one correspondence with with words of length n 2 (just by adding a Y Z onto the end, which we can always do.) So there are f n 2 words of this sort. Words that don t end in Y Z end in either X, or ZZ. In either case, we can remove the last letter to get a word of length n 1; conversely, any word of length n 1 can have a X or Z stuck on the end. So there are 2f n 1 words of this type. It follows that, for n 2, f n 2f n 1 + f n 2, and f 0 1, f 1 2. To solve the recurrence, let F (t) n 0 f nt n. Then so F (t) 1/(1 2t t 2 ), as before. F (t) f 0 + t(f 1 2f 0 ) + 2tF (t) + t 2 F (t) 1 + 2tF (t) + t 2 F (t) Factor: 1 2t t 2 (1 (1 + 2)t)(1 (1 2)t). Then, expanding by partial fractions, F (t) (1/4)(2 2) 1 (1 + 2t) + (1/4)(2 + 2) 1 (1 2t), so f n 1 4 (2 2)(1 + 2) n (2 + 2)(1 2) n, for n Let g n be the number of ways to select a breakfast consisting of an odd number of eggs, at most four pieces of toast, and at least one slice of bacon. Find an

6 ordinary generating function for the sequence {g n } and use it to compute g n in terms of n. Solution: A breakfast is completely determined by integers a 1, a 2, and a 3, where 0 < a 1 is odd, 0 a 2 4, and a 3 1. By the product principle for ordinary generating functions, g n t n (t + t 3 + t 5 + )(1 + t + t 2 + t 3 + t 4 )(t + t 2 + t 3 + ) n 0 t 1 t 5 t 1 t 2 1 t 1 t t2 (t + t 2 + t 3 + t 4 + t 5 ) (1 t) 2 (1 + t) 6 + 4t + 2t 2 + t 3 + 9t2 + 2t 6 (1 t) 2 by long division (1 + t) 6 + 4t + 2t 2 + t (1 t) t t 6 + 4t + 2t 2 + t 3 ( + 5/2(n + 1) 35/4 + 1/4( 1) n )t n, n0 from which we see g n (10n 45 + ( 1) n )/4 for n 4, and g 0 g 1 0, g 2 1, g Show that the Stirling numbers satisfy, for each k 1, the generating function identity S(n, k)t n t k (1 t)(1 2t) (1 kt). n How many different ways are there to label each face of a cube with an arrow, taking rotational symmetry into account? [Use Burnside s Lemma.] Solution: Let X be the set of labellings, without regard to symmetry, and let G be the symmetries of the cube, permuting the elements of X. We need to compute the number of labellings fixed by each symmetry in G. g 1: every labelling is fixed by the identity permutation, of course. Each face can be labelled in four ways, and there are six faces, so F 1 X and X 4 6. g a 90 rotation around a face axis: the arrow on a face that contains the axis of symmetry can t be sent to itself. So F g. The same argument applies to the 180 rotations around a face axis. If g is a vertex-rotation, recall g has two 3-cycles of faces that share a vertex. If we choose an arrow for one of those faces, there is only one way to chose the other two so that the arrows are sent to each other by the 120 rotation. There are 2 three-cycles, so for each such g, we have F g 4 4.

7 if g is an edge-rotation, the faces fall into three two-cycles. For each, choosing an arrow on one face determines the arrow for the other face. [Picture?] So F g 4 3. Adding it all up, we find there are distinct labellings. 1 G g G F g 1 ( ) Show that the exponential generating function for the number of derangements, D n t n /n!, equals n0 e t 1 t, if we agree D 0 1. [For maximum nutritional value, try using the product principle, or the recurrence relation D n (n 1)(D n 1 + D n 2 ).] Solution: The fastest way is remember that 1/(1 t) n 0 n!tn /n! is the exponential generating function for the number of permutations of length n. Let d(t) n 0 D nt n /n!. Let s count the permutations by noticing that every permutation has some subset S of fixed points, and the restriction of the permutation to the remaining elements [n] S is a derangement. The number of ways to let a set S be the fixed points of a permutation is just 1, for all k S 0. The e.g.f. for the sequence (1, 1, 1,...) is e t. By the product principle for exponential generating functions, every permutation is obtained by putting a derangement on a subset and making the complementary elements fixed points, so e t d(t) 1 1 t. That is, d(t) e t /(1 t), as required.