Given,

$${}_{90}^{232}Th\buildrel {} \over \longrightarrow {}_{82}^{212}pb$$

We know that $$\alpha$$-decay is given as

$${}_Z^AX\buildrel {} \over \longrightarrow {}_{Z - 2}^{A - 4}Y + {\alpha ^ - }$$ particle

and $$\beta$$-decay is given as

$${}_Z^AX\buildrel {} \over \longrightarrow {}_{Z + 1}^AX + {\beta ^ - }$$ particle

In above decay process change in mass number A is 20 and change in atomic number Z is 8.

Therefore, change in mass number A $$\Rightarrow$$ number of $$\alpha$$-particles emitted are

$${N_\alpha } = {{20} \over 4} = 5$$

Now, emission of 5 $$\alpha$$ particles implies atomic number reduces by 10 but change in atomic number is 8, therefore, number of $$\beta$$-particles emitted is 2.

Therefore, N_$$\alpha$$ = 5 and N_$$\beta$$ = 2

So, option (A) and option (C) are correct.