Math

1. Determine whether the relation R on the set {1, 2, 3, 4, 5} defined by (x, y) ∈ R if x + y ≤ 6 is reflexive, symmetric and/or transitive. 2. Give the equivalence class for the following: a. 3 modulo 5 b. 2 modulo 5 3. Give an algebraic proof for the binomial identity ( n k ) = ( n − 1 k − 1 ) + ( n − 1 k ). 4. A particular brand of shirt comes in 12 colors, has a male version and a female version, and comes in three sizes for each sex. How many different types of this shirt are made? 5. A multiple-choice test contains 10 questions. There are four possible answers for each question. a. In how many ways can a student answer the questions on the test if the student answers every question? b. In how many ways can a student answer the questions on the test if the student can leave answers blank? 6. How many positive integers between 5 and 31 a. are divisible by 3? Which integers are these? b. are divisible by 4? Which integers are these? c. are divisible by 3 and by 4? Which integers are these? 7. In how many ways can a photographer at a wedding arrange 6 people in a row from a group of 10 people, where the bride and the groom are among these 10 people, if a. the bride must be in the picture? b. both the bride and groom must be in the picture? c. exactly one of the bride and the groom is in the picture? d. the bride must be next to the groom? 8. How many bit strings contain exactly five 0s and 14 1s if every 0 must be immediately followed by two 1s? 9. How many bit strings of length 10 contain at least three 1s and at least three 0s? 10. What is the coefficient of x 101y 99 in the expansion of (2x − 3y) 200?

Step-by-step solution with explanation

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Step-by-step solution

Problem 1: Build the Relation R

We list all pairs where the sum is at most 6. This includes (1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1).

Problem 1: Check Reflexive, Symmetric, Transitive

(1,1),(2,2),(3,3) are in R but (4,4) needs 8≤6 (false) and (5,5) needs 10≤6 (false). So R is NOT reflexive. For symmetry: if x+y≤6 then y+x≤6, so YES symmetric. For transitivity: (3,3) and (3,1) are in R but check (4,2): 4+2=6≤6 ✓. Counterexample: (4,2)∈R and (2,4)∈R but (4,4): 8>6, so (4,4)∉R — NOT transitive.

Problem 2: Equivalence Classes mod 5

The equivalence class of 3 mod 5 is all integers that leave remainder 3 when divided by 5. Similarly, [2]_5 = {…,−8,−3,2,7,12,17,…} — all integers leaving remainder 2 when divided by 5.

Problem 3: Algebraic Proof of Binomial Identity

We start by writing each binomial coefficient using the factorial formula, then find a common denominator of k!(n-k)! and add.

Problem 3: Simplify to Get Left Side

Factor out (n-1)! from the numerator. The bracket k+(n-k)=n, so we get n·(n-1)!=n!, which equals C(n,k). This completes the algebraic proof.

Problems 4 & 5: Counting Shirts and Test Answers

Problem 4: Multiply choices — 12 colors × 2 sexes × 3 sizes = 72. Problem 5a: Each of 10 questions has 4 answers, all must be answered: 4^10 = 1,048,576 ways. Problem 5b: Each question now has 5 choices (4 answers + blank): 5^10 = 9,765,625 ways.

Problem 6: Integers Between 5 and 31 (exclusive)

Between 5 and 31 means 6 through 30. Divisible by 3: {6,9,12,15,18,21,24,27,30} — 9 integers. Divisible by 4: {8,12,16,20,24,28} — 6 integers. Divisible by both 3 and 4 (i.e., by 12): {12,24} — 2 integers.

Problem 7: Wedding Photographer Arrangements

(a) Fix bride in 6 possible spots, choose 5 more from 9 remaining and arrange: 6×P(9,5)=6×15120=90720. (b) Both bride and groom: choose 4 from 8, arrange all 6: P(8,4)×6×5=1680×30=50400. Wait — arrange 6 people in a row: fix bride in 6 spots, groom in 5 remaining spots, fill 4 spots from 8: 6×5×P(8,4)=30×1680=50400. (c) Exactly one of bride/groom: 2×[6×P(9,5) ways with one specific person] but subtract both-in cases: 2×90720−2×50400 = 181440−100800 = 80640. (d) Bride next to groom: treat them as a block of 2 (2 internal orders), arrange block+4 others chosen from 8 in 6-person row — choose 4 from 8: P(8,4) ways, place the couple-block as an adjacent unit in 5 possible consecutive positions in the row of 6, times 2: 2×5×P(8,4)=10×1680=16800.

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