Use counting principles to establish that we always have (when n is a non-negative integer while...

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Use counting principles to establish that we always have (when n is a non-negative integer while r and y are real numbers): (x + y)² = Σ ((^) ~*^*^*^*) k=0 Then use this identity to establish the following, by setting z = y = 1: = Σ() k=0 Use a combinatorial proof to show that when n and k are integers with n>k>0 we have: (²) - (n − k) = (k+ 1) - (x + 1) Let P(n. k) be the number of permutations of k elements taken from n elements. That is. P(n. k) counts how many ways we can take & distinct numbers from {1,2.....n) and then reorder them. Prove that when n and k are integers so that n>k>2 that: P(n. k) = P(n-1,k) + kP(n − 1. k − 1) (HINT: One way to do this is to think about choosing and reordering numbers from {1.2.....n} and dividing into cases based on whether you decide to pick 1 or not.) Use counting principles to establish that we always have (when n is a non-negative integer while r and y are real numbers): (x + y)² = Σ ((^) ~*^*^*^*) k=0 Then use this identity to establish the following, by setting z = y = 1: = Σ() k=0 Use a combinatorial proof to show that when n and k are integers with n>k>0 we have: (²) - (n − k) = (k+ 1) - (x + 1) Let P(n. k) be the number of permutations of k elements taken from n elements. That is. P(n. k) counts how many ways we can take & distinct numbers from {1,2.....n) and then reorder them. Prove that when n and k are integers so that n>k>2 that: P(n. k) = P(n-1,k) + kP(n − 1. k − 1) (HINT: One way to do this is to think about choosing and reordering numbers from {1.2.....n} and dividing into cases based on whether you decide to pick 1 or not.)

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To establish the identity 3 we can use the counting principle Lets break down the steps 1 Consider the lefthand side LHS of the equation 3 This represents choosing 3 elements from a set of z y element... View the full answer