Match Column – I with Column – II.  
 

	Column – I		Column – II
A)	The total number of three digit numbers, the sum of whose  Digits is even is equal to	P)	18
B)	Total number of positive integral solutions of the equation  $xyz=140$   is equal to	Q)	54
C)	Total number of positive integral solutions of  $x+y+z\le 10$   is equal to	R)	120
D)	If the cubic  $x^3+a{x}^2+bx+c$ is divisible by  $x^2+1,$ then   number of three digit numbers of the form  $abc$ or  $bca$  which can be formed is equal to	S)	450

A) Total number of three digit numbers =  $9\times 10\times 10.$
Half of which will have sum of digit even.
B)  $xyz=140=2^2\mathrm{.5.7}$
Therefore number of positive integral solutions =  $3\times 3{\times }^4{C}_2=54$ 

C)  $x+y+z+t=10,$ where   $1\le x,y,z\le 8,0\le t\le 7$

Therefore desired number of solutions = $coeff.of{x}^{10}$  in  ${(x+x^2+\mathrm{....}+x^8)}^3\times (1+x+x^2+\mathrm{....}+x^7){=}^{10}{C}_7{=}^{10}{C}_7{=}^{10}{C}_3=120$
D) We must have  $i^3+a{i}^2+bi+c=0$  and  ${(-i)}^3+a{(-i)}^2+b(-i)+c=0\Rightarrow b=1\&a=c$
Therefore number of numbers of type abc or cba is   ${ }^9{C}_1=9$

Number of numbers type bac or bca is  ${ }^{10}{C}_1=10$
But 111 is included in both the counting.