Answer: Option A. -> (2n)!n!
:
A
$\mathrm{1.3.5......}(2n-1)2^n$ = $\frac{\mathrm{1.2.3.4.5.6....}(2n-1)\left(2n\right)2^n}{\mathrm{2.4.5.....2}n}$
= $\frac{\left(2n\right)!2^n}{2^n\left(\mathrm{1.2.3......}n\right)}$ = $\frac{\left(2n\right)!}{n!}$

Answer: Option C. -> 9
:
C
Since number of derangements in such a problems is given by
$n!\{1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...............(-1{)}^n\frac{1}{n!}\}$
$\therefore$ Number of dearangements are
= $4!\{\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}\}$ = 12 -4 + 1 = 9

Answer: Option B. -> 810
:
B
The total number of numbers between 10 and 1000 are 989 but we have to form the numbers by using numerals 1,2,........9,so the numbers containing any '0' would be excluded i.e., Required number of ways
= 989 - $\left\{\begin{array}{c}20,30,40,...........................100=9\\ 101,102,...........................200=19\\ 201,...............................300=19\\ .........................................\\ .........................................\\ 901,..............................900=18\end{array}\right\}$
= $989-(9+18+19\times 8)$
Aliter:Between 10 and 1000, the numbers are of 2 digits and 3 digits.
Since repetition is allowed, so each digit can be filled in 9 ways.
Therefore number of 2 digit numbers = $989-(9+18+19\times 8)$
Aliter:Between 10 and 1000, the numbers are of 2 digits and 3 digits.
Since repetition is allowed, so each digit can be filled in 9 ways.
Therefore number of 2 digit numbers = $9\times 9$ = 81
and number of 3 digit numbers $9\times 9\times 9$ = 729
Hence total ways = 81 + 729 = 810.

Answer: Option B. -> 41
:
B
$\frac{56!}{(50-r)!}\times \frac{(51-r)!}{54!}$
= $\frac{30800}{1}\Rightarrow 56\times 55\times (51-r)$ = $30800\Rightarrow r$ = 41

Answer: Option A. -> 216
:
A
We know that a five digit number is divisible by 3, if and only if sum of its digits (= 15) is divisible by 3, therefore we should not use 0 or 3 while forming the five digit numbers.
Now,
(i) In case we do not use 0 the five digit number can be formed (from the digit 1, 2, 3, 4, 5) in ${}^5{P}_5$ways.
(ii) In case we do not use 3, the five digit number can be formed (from the digit 0, 1, 2, 4, 5) in ${}^5{P}_5-{}^4{P}_4$= 5! - 4!= 120 - 24 = 96 ways.
(${}^4{P}_4=$ cases when $0$ is at the first position)
$\therefore$The total number of such 5 digit number = ${}^5{P}_5+({}^5{P}_5-{}^4{P}_4)=120+96=216$

Answer: Option C. -> m!n!(2!)2(m−2)!(n−2)!
:
C
${}^m{C}_2.{}^n{C}_2$

Answer: Option D. -> 28
:
D
The number of straight lines is ${}^8{C}_2$ = 28

Answer: Option B. -> 600
:
B
The number of ways in which this can be done = 6! – 5! = 600

Answer: Option C. -> 210
:
C
The number of ways in which a man travel from Hyderabad to Triupati is 15 and back to Hyderabad is 14 and hence the total number of ways =15 $\times$ 14 = 210

Answer: Option C. -> 6!2!
:
C
The number of ways in which this can be done = 6! 2!