Let  $n\left(P\right)$ represents the number of points  $P(\alpha ,\beta )$ lying on the rectangular hyperbola $xy=15!$ , under the conditions given in column I, match the value of  $n\left(P\right)$ given in column II.

	Column –I		Column –II
(A)	$\alpha ,\beta \in I$	(p)	32
(B)	$\alpha ,\beta \in I^{+}andHCF(\alpha ,\beta )=1$	(q)	64
(C)	$\alpha ,\beta \in I^{+}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\alpha \text{\hspace{0.17em}\hspace{0.17em}divides\hspace{0.17em}\hspace{0.17em}}\beta$	(r)	96
(D)	$\alpha ,\beta \in I^{+}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}HCF(\alpha ,\beta )=35$	(s)	4032
		(t)	8064

 $xy=15!=2^{11}{3}^6{5}^3{7}^2{11}^1{13}^1$
(A) No. of the integral solutions = no. of ways of fixing x 
= the no. of factors of 15! 
 $=(1+11)(1+6)(1+3)(1+2)(1+1)(1+1)=4032$
$\Rightarrow$  Total no. of integral solutions   $=2\times 4032=8064$
(B)  $HCF(\alpha ,\beta )=1$. So identical primes should not be separated 
So, no. of solutions  $=2^6=64$
(C) The largest number whose perfect square can be made with $15!$  is  $2^5{3}^5{5}^{-1}{7}^1$
So the no. of ways of selecting x will be 
 $(1+5)(1+3)(1+1)(1+1)=96$
(D) Let  $\alpha =35{\alpha }_1$ and $\beta =35\beta$  where  $HCF({\alpha }_1,{\beta }_1)=1$
Now,  $\alpha \beta =15!\Rightarrow {\alpha }_1{\beta }_1=2^{11}{3}^6{5}^1{11}^1{13}^1$
So, no. of solutions  $=2^5=32$