ĐKXĐ: y>=3
TA có: \(A_{y}^3+C_{y}^{y-2}=14y\)
=>\(\frac{y!}{\left(y-3\right)!}+\frac{y!}{\left(y-y+2\right)!\cdot\left(y-2\right)!}=14y\)
=>\(\frac{y\left(y-1\right)\left(y-2\right)}{1}+\frac{y\left(y-1\right)}{2}=14y\)
=>\(\frac{2y\left(y-1\right)\left(y-2\right)+y\left(y-1\right)}{2}=14y\)
=>y(y-1)[2(y-2)+1]=28y
=>y(y-1)(2y-3)-28y=0
=>y[(y-1)(2y-3)-28]=0
=>(y-1)(2y-3)-28=0
=>\(2y^2-5y+3-28=0\)
=>\(2y^2-5y-25=0\)
=>\(2y^2-10y+5y-25=0\)
=>(y-5)(2y+5)=0
=>y=5(nhận) hoặc y=-5/2(loại)
\(M=A_{y+1}^4+3\cdot C_{y}^3\)
\(=A_{5+1}^4+3\cdot C_5^3=A_6^4+3\cdot C_5^3\)
\(=\frac{6!}{\left(6-4\right)!}+3\cdot\frac{5!}{\left(5-3\right)!\cdot3!}=\frac{6!}{2!}+3\cdot\frac{5\cdot4}{2!}=6\cdot5\cdot4\cdot3+3\cdot5\cdot2\)
=30*12+15*2
=360+30
=390
=>Chọn B
