How many ways can 12 boys be seated around a round table if two particular boys must sit together? 2. How many ways can six board of directors be seated on a circular table?

Welcome back Eniola — missed you this week. Here are the solutions to your assignment questions: 1. Seating 12 boys around a round table with 2 particular boys together Step 1: Treat the 2 particular boys who must sit together as a single unit. This reduces the number of entities to arrange to (12 - 2) + 1 = 11 units. Step 2: Calculate the number of ways to arrange these 11 units around a circular table. For circular permutations, the number of ways to arrange n distinct items is (n-1)!. So, for 11 units, the number of arrangements is (11-1)! = 10!. 10! = 3,628,800 Step 3: Account for the arrangements of the 2 particular boys within their unit. The 2 boys can sit in 2! ways within their unit (Boy A then Boy B, or Boy B then Boy A). 2! = 2 × 1 = 2 Step 4: Multiply the arrangements from Step 2 and Step 3 to get the total number of ways. Total ways = (arrangements of units) × (arrangements within the unit) 3,628,800 × 2 = 7,257,600 The 2 particular boys must sit next to each other in 7,257,600 ways. 2. Seating six board of directors on a circular table Step 1: Identify the number of people to be arranged. There are n=6 directors. Step 2: Apply the formula for circular permutations. The number of ways to arrange n distinct items in a circle is (n-1)!. (6-1)! = 5! Step 3: Calculate the factorial. 5! = 5 × 4 × 3 × 2 × 1 = 120 Six board of directors can sit on a circular table in 120 ways. 3. Expand the given expressions a) Expand (2x-3)^4 Step 1: Use the binomial theorem (a+b)^n = _k=0^n nk a^n-k b^k. Here, a=2x, b=-3, and n=4. Step 2: Calculate each term of the expansion. Term 1 (k=0): 40(2x)^4(-3)^0 = 1 · (16x^4) · 1 = 16x^4 Term 2 (k=1): 41(2x)^3(-3)^1 = 4 · (8x^3) · (-3) = -96x^3 Term 3 (k=2): 42(2x)^2(-3)^2 = 6 · (4x^2) · 9 = 216x^2 Term 4 (k=3): 43(2x)^1(-3)^3 = 4 · (2x) · (-27) = -216x Term 5 (k=4): 44(2x)^0(-3)^4 = 1 · 1 · 81 = 81 Step 3: Sum the terms to get the expanded form. (2x-3)^4 = 16x^4 - 96x^3 + 216x^2 - 216x + 81 b) Expand (x-2)^3 (Based on your notes for Question 2, where a=x, b=-2, A=3) Step 1: Use the binomial theorem (a+b)^n = _k=0^n nk a^n-k b^k. Here, a=x, b=-2, and n=3. Step 2: Calculate each term of the expansion. Term 1 (k=0): 30(x)^3(-2)^0 = 1 · x^3 · 1 = x^3 Term 2 (k=1): 31(x)^2(-2)^1 = 3 · x^2 · (-2) = -6x^2 Term 3 (k=2): 32(x)^1(-2)^2 = 3 · x · 4 = 12x Term 4 (k=3): 33(x)^0(-2)^3 = 1 · 1 · (-8) = -8 Step 3: Sum the terms to get the expanded form. (x-2)^3 = x^3 - 6x^2 + 12x - 8 Drop the next question! 📸