Question

Column I		Column II
A	Number of all six digit natural numbers such that the sum of their digits is 10 and each of the digits $0,1,2,3$ occurs atleast once in them is	P	350
B	A question paper consists of 2 parts A and B. Part-A has 4 questions with 1 alternative each and part-B has 3 question without any alternative. Number of ways in which one can select the question when atleast one question must be attempted from each part, is	Q	405
C	Number of ordered pairs of positive integers $(a,b)$ such that the least common multiple of ' $a$ ' and 'b' is $2^2\cdot 5^4\cdot 11^4$	R	490
		S	560

• A. (A) R; (B) S; (C) P
• B. (A) R; (B) S; (C) Q
• C. (A) R; (B) S; (C) R
• D. (A) R; (B) S; (C) S

Answer 1

Answer: (B)

(A) Six digits numbers with sum of digits 10
$0,1,2,3$ occurs at least once remaining two digits must give the sum 10 . Hence possible cases two digits if their sum is 4 , can be
$04,13,22$
six digit numbers with
(i) $001234=\frac{6!}{2!}-5$ ! $=240$
(ii) $012313=\frac{6!}{2!\cdot 2!}-\frac{5!}{2!\cdot 2!}=180-30=150$
(iii) $012322=\frac{6!}{3!}-\frac{5!}{3!}=120-20=100$
Total $=490$ Ans. $\Rightarrow$ (R)

(C) LCM of a and b is $2^2\cdot 5^4\cdot 11^t$
let $a=2^r\cdot 5^s\cdot t^4$ and $b=2^{r_1}\cdot 5^{s_1}\cdot 11^{t_1}$

|||$ly$ and $t_1$ can be taken in 9 ways. Total ways $=5\cdot 9\cdot 9=405$ Ans. $\Rightarrow (Q)$