Let's solve each question **step by step** with explanations and conclusions: --- ## **Question 1** ### **How many bit strings of length 9 contain...** #### **a. exactly 5 ones?** **Step 1:** A bit string of length 9 has 9 positions. **Step 2:** We must select exactly 5 of those 9 positions to be ones (the rest will be zeros). **Step 3:** The number of ways to choose 5 positions from 9 is denoted as a binomial coefficient: \[ \text{Number of ways} = \binom{9}{5} \] **Step 4:** Calculate: \[ \binom{9}{5} = \frac{9!}{5! \cdot 4!} = \frac{362880}{120 \cdot 24} = \frac{362880}{2880} = 126 \] **Conclusion:** **126 bit strings** of length 9 contain exactly 5 ones. --- #### **b. at most 5 ones?** **Step 1:** "At most 5 ones" means 0, 1, 2, 3, 4, or 5 ones. **Step 2:** Add binomial coefficients for all these cases: \[ \sum_{k=0}^5 \binom{9}{k} \] **Step 3:** Calculate each: - \(\binom{9}{0} = 1\) - \(\binom{9}{1} = 9\) - \(\binom{9}{2} = 36\) - \(\binom{9}{3} = 84\) - \(\binom{9}{4} = 126\) - \(\binom{9}{5} = 126\) Sum: \(1 + 9 + 36 + 84 + 126 + 126 = 382\) **Conclusion:** **382 bit strings** of length 9 contain at most 5 ones. --- #### **c. at least 5 ones?** **Step 1:** "At least 5 ones" means 5, 6, 7, 8, or 9 ones. **Step 2:** Add binomial coefficients for all these cases: \[ \sum_{k=5}^9 \binom{9}{k} \] **Step 3:** Since \(\binom{9}{k} = \binom{9}{9-k}\), you can use symmetry or calculate directly: - \(\binom{9}{5} = 126\) - \(\binom{9}{6} = 84\) - \(\binom{9}{7} = 36\) - \(\binom{9}{8} = 9\) - \(\binom{9}{9} = 1\) Sum: \(126 + 84 + 36 + 9 + 1 = 256\) **Conclusion:** **256 bit strings** of length 9 contain at least 5 ones. --- ## **Question 2** ### **How many functions are there from a set with 9 elements to a set with 42 elements?** **Step 1:** Each element in the domain (9 elements) can be mapped to any of the 42 elements in the codomain. **Step 2:** For each element in the domain, there are 42 choices, so the total number of functions is: \[ 42^9 \] **Conclusion:** There are **42⁹** functions from a set with 9 elements to a set with 42 elements. --- ## **Question 3** Let \( X = \{24, 42, 57\} \). Define \( g : X \rightarrow X \) as \( g(24) = 42, g(42) = 57, g(57) = 42 \). Define \( f : X \rightarrow X \) as \( f(24) = 57, f(42) = 24, f(57) = 42 \). ### **What is \( f \circ g \)?** This means \( f(g(x)) \): - \( f(g(24)) = f(42) = 24 \) - \( f(g(42)) = f(57) = 42 \) - \( f(g(57)) = f(42) = 24 \) So, - 24 → 24 - 42 → 42 - 57 → 24 ### **What is \( g \circ f \)?** This means \( g(f(x)) \): - \( g(f(24)) = g(57) = 42 \) - \( g(f(42)) = g(24) = 42 \) - \( g(f(57)) = g(42) = 57 \) So, - 24 → 42 - 42 → 42 - 57 → 57 --- ## **Question 4** Given \( P(k) \implies P(k+3) \) for every positive integer \( k \). To prove \( P(n) \) is true for all positive integers \( n \), you need to prove it for **three consecutive values** because every number can be written as \( 3m, 3m+1, 3m+2 \). **Conclusion:** You need to prove: \[ \boxed{P(1) \text{ and } P(2) \text{ and } P(3)} \] --- ## **Question 5** ### **Proof by Induction: Number of subsets** - The subsets of \(\{1\}\): empty set and \(\{1\}\), so \( a_1 = 2 \). - For any \( k > 1 \), each subset of \(\{1,2,...,k\}\) either contains \( k \) or it doesn't. - For each subset of \(\{1,2,...,k-1\}\), you can form two subsets of \(\{1,2,...,k\}\) (with or without \( k \)). - Thus, \( a_k = 2a_{k-1} \). - The recurrence: \( a_n = 2a_{n-1} \). --- ### **Summary Table for Answers:** | **Q#** | **Answer** | |-------|--------------------------------------------------| | 1a | 126 | | 1b | 382 | | 1c | 256 | | 2 | \( 42^9 \) | | 3 | \( f \circ g: \) 24→24, 42→42, 57→24 | | | \( g \circ f: \) 24→42, 42→42, 57→57 | | 4 | \( P(1) \) and \( P(2) \) and \( P(3) \) | | 5 | \( a_1 = 2 \), \( a_k = 2a_{k-1} \), \( a_n = 2a_{n-1} \) | Let me know if you need any additional explanation for any step!

Let's analyze the problem step by step: --- ## **Understanding the problem** The question asks: **"What is the percentage chance that their offspring will be XY with normal colour vision?"** Assuming this refers to a typical genetics problem involving sex-linked inheritance of colour vision: - **Males** are XY chromosomes. - **Females** are XX chromosomes. - The gene for colour vision is **X-linked**. - The allele for **normal colour vision** is **C**. - The allele for **colour blindness** is **c**. - **Normal colour vision**: **XC** (dominant) or **XC XC** (homozygous dominant). - **Colour blindness**: **c c** (homozygous recessive). --- ## **Step 1: Determine the parents' genotypes** Given it's a couple, but the problem doesn't specify their individual traits; based on typical questions, common assumptions are: - The **mother** is **carrier**: **XC Xc** (heterozygous, normal vision but carrier). - The **father** is **normal vision**: **XC Y** (since males have only one X, the allele on X determines vision). --- ## **Step 2: Possible offspring genotypes** ### **For male offspring (XY):** - They inherit **Y** from the father. - They inherit **X** from the mother, which can be either **XC** or **Xc**. **Possible male genotypes:** - **XC Y**: normal vision. - **Xc Y**: colour blind. ### **Probability for male offspring:** - The mother can pass **XC** or **Xc** with equal probability (50% each). - Since the father is **Y**, the male genotype depends solely on the mother's X. So, the chance: - **Normal vision males (XC Y):** 50% - **Colour blind males (Xc Y):** 50% --- ### **For female offspring (XX):** - They inherit **X** from both parents. - Mother: **XC Xc** (carrier) - Father: **XC** **Possible female genotypes:** - **XC X** (normal vision, non-carrier) - **XC Xc** (carrier, normal vision) **Probabilities:** - **XC X:** 50% - **XC Xc:** 50% **Note:** Females need **two copies of recessive** (Xc) to be colour blind, so: - All females will have **normal vision** (either **XC X** or **XC Xc**). --- ## **Step 3: Focus on XY offspring with normal vision** The question asks **"chance that their offspring will be XY with normal colour vision"**. - As shown, **male** offspring are **XY**. - The probability that a male is **XY with normal vision** is **50%** (from Step 2). --- ## **Step 4: Convert probability to percentage** - The probability that their **offspring** (specifically, a **male XY**) will have **normal colour vision** is **50%**. --- ## **Final answer:** **Percentage chance = 50** --- ## **Summary:** | Step | Explanation | Probability | |-----------------------------------|-------------------------------------------------------|--------------| | Male offspring with normal vision | Inherit **XC** from mother, **Y** from father | 50% | | Female offspring with normal vision | All females have normal vision (either **XC X** or **XC Xc**) | 100% (but not XY) | | **XY with normal vision** | Only male XY offspring with **XC Y** | **50%** | --- **Therefore, the percentage chance that their offspring will be XY with normal colour vision is:** **50** --- **(Enter: 50)**