The formula for permutations is ^kP_m = (k!)/((k-m)!).

Okay, let's solve problem 2(i). We need to find n given the equation ^nP_4 = 20 · ^nP_2. The formula for permutations is ^kP_m = (k!)/((k-m)!). Step 1: Write out the permutation terms using the formula. ^nP_4 = (n!)/((n-4)!) ^nP_2 = (n!)/((n-2)!) Step 2: Substitute these expressions into the given equation. (n!)/((n-4)!) = 20 · (n!)/((n-2)!) Step 3: Simplify the equation. Since n! is on both sides and n 4 (for ^nP_4 to be defined), n! ≠ 0. We can divide both sides by n!. (1)/((n-4)!) = (20)/((n-2)!) Step 4: Express (n-2)! in terms of (n-4)!. We know that (n-2)! = (n-2)(n-3)(n-4)!. Step 5: Substitute this into the equation. (1)/((n-4)!) = (20)/((n-2)(n-3)(n-4)!) Step 6: Cancel (n-4)! from both sides. 1 = (20)/((n-2)(n-3)) Step 7: Solve for n. (n-2)(n-3) = 20 n^2 - 3n - 2n + 6 = 20 n^2 - 5n + 6 = 20 n^2 - 5n - 14 = 0 Step 8: Solve the quadratic equation by factoring. We look for two numbers that multiply to -14 and add to -5. These numbers are -7 and 2. (n-7)(n+2) = 0 This gives two possible solutions: n-7 = 0 n = 7 n+2 = 0 n = -2 Step 9: Check for valid solutions. For ^nP_r to be defined, n must be a non-negative integer and n r. In this case, n 4. The value n=7 satisfies n 4. The value n=-2 does not satisfy n 4. Therefore, the only valid solution is n=7. The value of n is 7. That's 2 down. 3 left today — send the next one.