Download CMSC 203 Spring 2011 Exam 3: Combinatorics, Equivalence Relations, Probability and more Exams Discrete Structures and Graph Theory in PDF only on Docsity!
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1. (6 points) How many license plates can a state produce if the plates can contain 7 characters (from 26 let- ters and 10 digits) if a certain pair of letters cannot be adjacent to one another and all the characters must be distinct?
ALL - (Plates with the pair as XY) - (Plates with the pair as YX)
= P(36,7) - P(35,6) - P (35,6) = (36! / 29!) - 2(35! / 28!)
2. (6 points) How many ways can a teacher choose 10 students from a class of 13 Boys and 16 Girls, if she must choose the same number of boys and girls?
(Choose 5 boys) AND (Choose 5 girls) = C(13,5)C(16,5) = (13! / 8!5!)(16! / 11!5!) or (13!16!)/(8!5!11!5!)
3. (6 points) How many orderings are there of the letters of the word ELECTRICALENGINEERING?
ELECTRICALENGINEERING = EEEEELLCCTRRIIIANNNGG
Orderings = 21! / (5!2!2!2!3!3!2!) or C(21,5)C(16,2)C(14,2)C(12,1)C(11,2)C(9,3)C(6,1)C(5,3)C(2,2)
4. (6 points) How many ways can I seat 12 people around a circular table?
Orderings = (12 - 1)! = 11!
5. (6 points) How many ways can I fill a box of 50 chocolates from 15 types if I must have at least 2 of each type in the box?
Slots = 50, Categories = 15, Transitions = 15 - 1 = 14, Total Restrictions = 2(15) = 30 so
Total Boxes = C(50 + 14 - 30 , 50 - 30) = C(34, 20) = C(34, 14) = 34! / 20!14!
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Let R be the relation on Z given by R = {( a,b ) | a,b ∈ Z and a ≡ b mod 10}.
6. (6 points) Show the R is Reflexive.
Let x be an Integer, then ( x − x ) = 0 = 10(0), and 0 is an Integer, hence ( x , x ) is in R.
Therefore, R is Reflexive.
7. (6 points) Show the R is Symmetric.
Let x and y be Integers with ( x, y ) in R. This implies ( x − y ) = 10 k , for some Integer k.
Thus, ( y − x ) = −( x − y ) = − 10 k = 10(− k ). Since k is an Integer, (− k ) is an Integer, hence ( y, x ) is in R.
Therefore, R is Symmetric.
8. (6 points) Show the R is Transitive.
Let x , y , and z be Integers with ( x, y ) and ( y, z ) in R. Thus ( x − y ) = 10 k and ( y − z ) = 10 m for some Integers k and m.
This yields ( x − z ) = ( x − y ) + ( y − z ) = 10 k + 10 m = 10( k + m ). Since k and m are Integers, ( k + m ) is also, hence ( x, z ) is in R.
Therefore, R is Transitive.
9. (6 points) Describe the partition of Z induced by R.
Partition( Z ) = { [0], [1], [2], [3], [4], [5], [6], [7], [8], [9] }
13. (6 points) (a) For a collection of 45 coins, if 27 are quarters, 12 are quarters from the 1990’s, and 20 are coins from the 1990’s. Show that the probability the a coin being a quarter is INDEPENDENT from it being from the 1990’s?
Denote E = {Quarters} and F = {Coins from the 1990’s}
Test 1: Show P(E | F) = P(E). Since P(E | F) = |E ∩ F| / |F| = 12/20 = 3/5 = 27/45 = P(E), we see that E is Independent of F. Test 2: North = 12, South = (45 - 27 - 20 + 12) = 10, East = (27 - 12) = 15 and West = (20 - 12) = 8, so North(South) = 12(10) = 120 = 15(8) = East(West), so E is Independent of F.
14. (6 points) Find the probability of rolling 2 fair dice and getting a total of at least 10 if the first die rolls at least 4?
First Roll Second Roll(s) 4 6 5 5, 6 6 4, 5, 6 F =
E = {Sum > 10} and F = {First = 4, 5, or 6} then |E ∩ F| = 6 and |F| = 18, E = therefore P(E | F) = 6/18 = 1/3.
15. (6 points) (a) Find the Disjunctive Normal Form for the Boolean Polynomial F( w,x,y,z ) = w’z + w x’y’
F( w,x,y,z ) = w’z + w x’y’ = w’ ( x + x’ )( y + y’ ) z + w x’y’ ( z + z’ )
= w’xyz + w’x’yz + w’xy’z + w’x’y’z + wx’y’z + wx’y’z’
16. (6 points) Find the Disjunctive Normal Form of the polynomial with Truth Table:
x y z F( x,y,z ) 1 1 1 0 1 1 0 1 >>> xyz ’ 1 0 1 0 1 0 0 1 >>> xy’z ’ 0 1 1 0 0 1 0 0 0 0 1 1 >>> x’y’z so F( x,y,z) = xyz ’ + xy’z ’ + x’y’z 0 0 0 0
