This can be solved by Newton's binomial formula
Tk+1 = (n k) a∧(n-k) b∧k k+1=7 => k=6, n=11 , a=-3x and b=-2y
T7= (11 6) (-3x)∧(11-6) (-2y)∧6 = (11*10*9*8*7)/(1*2*3*4*5) (-3x)∧5 (-2y)∧6
T7= 462 (-3x)∧5 (-2y)∧6
The correct answer is A.
Good luck!!!
[tex]\displaystyle T_r=\binom{n}{r-1}x^{n-r+1}y^{r-1}\\\\\\T_7=\binom{11}{6}(-3x)^{11-7+1}(-2y)^{7-1}\\\\T_7=\dfrac{11!}{6!5!}(-3x)^5(-2y)^6\\\\T_7=\dfrac{7\cdot8\cdot9\cdot10\cdot11}{2\cdot3\cdot4\cdot5}(-3x)^5(-2y)^6\\\\T_7=462(-3x)^5(-2y)^6[/tex]
