What is the conductivity of a semiconductor sample having . . .

Question : What is the conductivity of a semiconductor sample having electron concentration of 5×10¹⁸ m^-3, hole concentration of 5×10¹⁹ m^-3, electron mobility of 2.0 m²V^-1s^-1 and hole mobility of 0.01 m²V^-1s^-1? (Take charge if electron as 1.6×10^-19C)

Doubt by Parish

Solution : 

We know, 
j=σE

I/A=σE
I=σEA
neVdA=σEA [∵I=neVdA]
neVd=σE
ne[Vd/E]=σ
neµ=σ [∵µ=Vd/E]

σ=enµ

where
σ = Conductivity

n = charge density 
µ = mobility of charge carriers

Since semiconductors have both electrons and holes this is why for Semiconductors

σ=eµ_en_e+eµ_hn_h)
σ=e(µ_en_e+µ_hn_h)
n_e =5×10¹⁸ m^-3 

n_h = 5×10¹⁹ m^-3 

µ_e = 2.0 m²V^-1s^-1 

µ_h = 0.01 m²V^-1s^-1 

e = 1.6×10^-19C 

σ=e(µ_en_e+µ_hn_h)

σ = 1.6×10^-19 (2×5×10¹⁸ +0.01×5×10¹⁹) 
σ = 1.6×10^-19 (10¹⁹+0.05×10¹⁹)
σ = 1.6 ×10^-19×10¹⁹(1+0.05)
σ = 1.6 (1.05)
σ = 1.68 Ω^-1m^-1