[tex]1)\,\,\,\frac{A_n^{4}*P_{n-4}}{P_{n-2}}} =42C_{\boxed{ }}^{5} \\ \\ \frac{A_n^{4}*(1*2*3*...*(n-4))}{1*2*3*.....*(n-4)(n-3)(n-2)} =42C_{\boxed{ }}^{5} \\ \\ \frac{A_n^{4}}{(n-3)(n-2)} =42C_{\boxed{ }}^{5} \\ \\ \frac{n(n-1)(n-2)(n-3)}{(n-3)(n-2)} =42C_{\boxed{ }}^{5} \\ \\ n(n-1)=42C_{\boxed{ 5}}^{5} \\ n(n-1) = 42 \\ => n=7[/tex]
=> Egalitatea este adevarata daca in patratel avem numarul 5 si n = 7
[tex]2)\,\,\, A_{3n-5}^{ n^{2}-2n}[/tex] este definit pentru n ∈ {2; 3; 5}
Conditiile sunt:
3n - 5 > 0 => valorile: {1; 4; 10}
n² - 2n ≥ 0 => valorile {0; 3; 15}
3n - 5 > n² - 2n => FALS deoarece:
penteu n = 5, 3n - 5 < n² - 2n
3)
x + 1 - (x - 1) = x + 1 -x + 1 = 2
[tex]A_{x+1}^{x-1}+14P_{x-1} \leq 30P_x \\ \\ \frac{P_{x+1}}{2} +14P_{x-1} \leq 30P_x \\ P_{x+1}+28P_{x-1} \leq 60P_x \\ P_{x+1}+28P_{x-1} - 60P_x \leq 0 \\ P_{x-1}( x(x+1)+28 - 60x) \leq 0 \\ x^{2} +x-60x+28 \leq 0 \\ x^{2} -59x+28 \leq 0 \\ \Delta = 59^{2} - 4*28 = 3369 \\ \sqrt{3369}=58,04 \\ => 1 \leq x \leq 58 [/tex]
[tex]4)\,\,\, P_{10} - P_{9} [/tex]
[tex]6)\,\,\, n = C_{14}^{3}*C_{18}^{5} [/tex]
