Download Exam Solutions for Math 213 - Problem Set 1 and more Exams Discrete Mathematics in PDF only on Docsity!
Math 213 Exam 1 (Solutions)
Prof. I.Kapovich March 6, 2007
Problem 1. Using induction, prove that for every integer n ≥ 1 we have:
n^2
n
Solution. We will prove that inequality (†) holds for every n ≥ 1 by induction on n. (1) For n = 1 the left hand side of (†) is 112 = 1 and the right-hand side of (†) is 2 − 11 = 2 − 1 = 1. Since 1 ≤ 1, the statement has been verified for n = 1. (2) Suppose now that n ≥ 1 and that (†) has been established for n. We need to show that it holds for n + 1, that is we need to prove
n^2
(n + 1)^2
n + 1
Since we know that for n inequality (†) holds, we add (^) (n+1)^12 to both
sides of (†) and get
n^2
(n + 1)^2
n
(n + 1)^2
If we can now show that
n
(n + 1)^2
n + 1
this will imply that inequality (∗) holds as required. We have: 1
n
(n + 1)^2
n + 1
n
(n + 1)^2
n + 1
n
(n + 1)^2
n + 1
(n + 1)^2 − n − n(n + 1) n(n + 1)^2
n^2 + 2n + 1 − n − n^2 − n n(n + 1)^2
n(n + 1)^2
The last inequality in this sequence is obviously true since n ≥ 1. Thus (∗ ∗ ∗) holds and therefore (∗) also holds, which completes the inductive step.
Problem 2. How many different strings can be made by permuting the letters of the word ILLINOIS if the two L’s must be consecutive?
Solution. We will apply the formula for the number of permutations with in- distinguishable objects. Since the two L’s must be consecutive, we consider them to be a single letter LL. Then the word ILLIN OIS has n = 7 letters: 3 I’s, 1 LL, 1 N , 1 O and 1 S. Hence the numbers of ways to permute letters of the word ILLINOIS if the two L’s must be consecutive is:
7! 3! 1! 1! 1! 1!
Problem 3. (a) Find the precise value of
i=0 C(100, i)(−2) i.
(b) Find the coefficient at x^6 in the expansion of (3 + x)^75. Simplify the answer to the extent possible.
Solution.
where there are C(4, 1) to choose a suit, C(13, 3) ways to pick a 3-card hand from that suit and C(49, 1) ways to “pick the fourth card out of the remaining 52 − 3 = 49 cards”. The problem with this argument is that every 4-card hand where all 4 cards are of the same suit is counted in formula (‡) more than once. Specifically, each such hand is counted in the above formula 4 times corresponding to 4 possible ways of splitting this hand into a 3-card hand and a 1-card hand. Observe, every hand with 3 cards of the same suit and the fourth card of a different suit is counted in (‡) once. The overall count in (‡), however, is incorrect.
Problem 5. How many integers n, where 1 ≤ n ≤ 1000, are divisible by 4 or 13? Solution. Let A be the set of all integers n, where 1 ≤ n ≤ 1000, that are divisible by 4 and let B be the set of all integers n, where 1 ≤ n ≤ 1000, that are divisible by 13. Then A∩B is the set of all in integers n, where 1 ≤ n ≤ 1000, that are divisible by 4·13 = 52. We need to find |A∪B|. Since 1000 = 4 · 250, 1000 = 13 · 76 .9230 and 1000 = 52 · 19 .23, we have |A| = 250, |B| = 76 and |A ∩ B| = 19. Therefore
|A ∪ B| = |A| + |B| − |A ∩ B| = 250 + 76 − 19 = 307.
