This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Right yo, let's go. To prove the identity ^nP_r = ^n-1P_r + r · ^n-1P_r-1, we will start with the right-hand side (RHS) and transform it into the left-hand side (LHS) using the definition of permutations, ^kP_m = (k!)/((k-m)!). Step 1: Write out the terms on the RHS using the permutation definition. ^n-1P_r = ((n-1)!)/(((n-1)-r)!) = ((n-1)!)/((n-r-1)!) ^n-1P_r-1 = ((n-1)!)/(((n-1)-(r-1))!) = ((n-1)!)/((n-1-r+1)!) = ((n-1)!)/((n-r)!) Step 2: Substitute these expressions back into the RHS. RHS = ((n-1)!)/((n-r-1)!) + r · ((n-1)!)/((n-r)!) Step 3: Find a common denominator for the two terms. We know that (n-r)! = (n-r) · (n-r-1)!. Multiply the first term by (n-r)/(n-r) to get a common denominator of (n-r)!. RHS = ((n-1)!)/((n-r-1)!) · ((n-r))/((n-r)) + (r(n-1)!)/((n-r)!) RHS = ((n-r)(n-1)!)/((n-r)!) + (r(n-1)!)/((n-r)!) Step 4: Combine the terms over the common denominator. RHS = ((n-r)(n-1)! + r(n-1)!)/((n-r)!) Step 5: Factor out (n-1)! from the numerator. RHS = ((n-1)! · ((n-r) + r))/((n-r)!) RHS = ((n-1)! · (n-r+r))/((n-r)!) RHS = ((n-1)! · n)/((n-r)!) Step 6: Rearrange the numerator to form n!. Since n! = n · (n-1)!, we can rewrite the numerator. RHS = (n!)/((n-r)!) Step 7: Recognize this as the definition of ^nP_r. RHS = ^nP_r This matches the LHS. Therefore, the identity is proven: ^nP_r = ^n-1P_r + r · ^n-1P_r-1 Send me the next one 📸