Question

sa se calculeze A$<var>A^3_5 - 4C^2_6</var>$

Answer 1

$A_5^3-4\cdot C_6^2=5\cdot4\cdot3-4\cdot\dfrac{6\cdot5}{1\cdot2}=60-60=0$

Am folosit formula $A_n^k=n(n-1)(n-2)...(n-k+1)$ (se retine si se aplica usor, pentru ca in partea stanga sunt exact k numere.)

Si $C_n^k=\dfrac{n(n-1)(n-2)...(n-k+1)}{k!}$