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ad a) {n \choose {n - 2}} - {n \choose 1} = 0 {n! \over (n - 2)! \cdot (n - (n - 2))!} - {n! \over 1! \cdot (n - 1)!} = 0 {n! \over (n - 2)! \cdot 2!} - {n! \over 1! \cdot (n - 1)!} = 0 {(n - 2)! \cdot (n - 1) \cdot n \over (n - 2)! \cdot 2} - {(n - 1)! \cdot n \over (n - 1)!} = 0 {n(n - 1) \over 2} - n = 0 n^2 - n - 2n = 0 n^2 - 3n = 0 n(n - 3) = 0 n = 0 lub n = 3 Zatem n = 3 ad b) {n \choose 2} - {n \choose 3} = 0 {n! \over 2! \cdot (n - 2)!} - {n! \over 3! \cdot (n - 3)!} = 0 {(n - 2)!(n - 1)n \over 2! \cdot (n - 2)!} - {(n - 3)!(n - 2)(n - 1)n \over 3! \cdot (n - 3)!} = 0 {(n - 1)n \over 2!} - {(n - 2)(n - 1)n \over 3!} = 0 {n^2 - n \over 2} - {n^3 - n^2 - 2n^2 + 2n \over 2 \cdot 3} = 0 {3n^2 - 3n \over 6} - {n^3 - n^2 - 2n^2 + 2n \over 6} = 0 {3n^2 - 3n - n^3 + n^2 + 2n^2 - 2n \over 6} = 0 {-n^3 + 6n^2 - 5n \over 6} = 0 -n^3 + 6n^2 - 5n = 0 -n(n^2 - 6n + 5) = 0 -n(n^2 - 5n - n + 5) = 0 -n(n(n - 5) - (n - 5)) = 0 -n(n - 1)(n - 5) = 0 n = 0 lub n = 1 lub n = 5 Zatem n = 5