1. **Problem statement:** Find the number of bytes (8-bit sequences) and answer related questions about bytes starting or ending with specific bit patterns. 2. **Definitions and formulas:** - A byte is a sequence of 8 bits, each bit can be 0 or 1. - Total number of bytes is $2^8$ because each bit has 2 choices. - For bytes starting or ending with specific bits, fix those bits and vary the rest. 3. **Solution:** **a) Number of bytes:** Each of the 8 bits can be 0 or 1 independently. $$\text{Total bytes} = 2^8 = 256$$ **b) Number of bytes that begin with 10 and end with 10:** - First two bits fixed as 1 and 0. - Last two bits fixed as 1 and 0. - Remaining bits: $8 - 4 = 4$ bits free. $$\text{Number} = 2^4 = 16$$ **c) Number of bytes that begin with 10 and do not end with 10:** - Total bytes beginning with 10: fix first two bits, free bits = 6. $$2^6 = 64$$ - Bytes beginning with 10 and ending with 10 (from b): 16 - So bytes beginning with 10 and NOT ending with 10: $$64 - 16 = 48$$ **d) Number of bytes that begin with 10 or end with 10:** - Bytes beginning with 10: $2^6 = 64$ - Bytes ending with 10: fix last two bits, free bits = 6, so $2^6 = 64$ - Bytes beginning with 10 and ending with 10 (intersection): 16 - Use inclusion-exclusion: $$64 + 64 - 16 = 112$$ --- 4. **Problem:** Find the number of 4-digit odd numbers with no repeated digits. - 4-digit number: digits $d_1 d_2 d_3 d_4$ with $d_1 \neq 0$. - Number is odd: last digit $d_4$ is odd: choices are 1,3,5,7,9 (5 choices). - No repeated digits. Step-by-step: - Choose $d_4$ (last digit): 5 choices. - Choose $d_1$ (first digit): cannot be 0 or $d_4$, so 8 choices. - Choose $d_2$: cannot be $d_1$ or $d_4$, so 8 choices. - Choose $d_3$: cannot be $d_1$, $d_2$, or $d_4$, so 7 choices. Total: $$5 \times 8 \times 8 \times 7 = 2240$$ --- 5. **Problem:** A discrete math exam has parts A, B, C each with 5 questions. How many ways to answer 7 questions selecting at least 2 from each part? - Let $a,b,c$ be number of questions chosen from parts A,B,C. - Constraints: $a+b+c=7$, $a,b,c \geq 2$. Possible $(a,b,c)$: - (3,2,2), (2,3,2), (2,2,3) Calculate number of ways for each: - For (3,2,2): choose 3 from 5 in A: $\binom{5}{3}=10$ - choose 2 from 5 in B: $\binom{5}{2}=10$ - choose 2 from 5 in C: $\binom{5}{2}=10$ - Total: $10 \times 10 \times 10 = 1000$ Similarly for (2,3,2) and (2,2,3), each also 1000 ways. Sum all: $$1000 + 1000 + 1000 = 3000$$ --- **Final answers:** - a) 256 - b) 16 - c) 48 - d) 112 - Number of 4-digit odd numbers with no repeated digits: 2240 - Number of ways to answer 7 questions with at least 2 from each part: 3000