Answer: Option C. -> 9.9!
:
C
Required number numbers = total - the number of numbers begining with 0 = $10!–9!=9.9!$

Answer: Option D. -> 24
:
D
By fundamental theorem of Multiplication = 4 $\times$ 3 $\times$ 2

Answer: Option A. -> 2240
:
A
The number of four digit numbers which satisfy the above condition = $8\times 8\times 7\times 5=2240$

Answer: Option D. -> 156
:
D
If 0 is in units place no. of ways = ${}^5{\text{P}}_3=60$
If 2 or 8 is in units place no. of ways = $2{(}^5{\text{P}}_3{-}^4{\text{P}}_2)=96$
Total : 60 + 96 = 156

Answer: Option A. -> n(nr−1)n−1)
:
A
$n+n^2+\cdots \cdots \cdots +n^r=\frac{n(n^r-1)}{n-1}$

Answer: Option A. -> 20
:
A
A student can enter the room in 5 ways but he can leave the room in 4 ways.
The total number of ways in which this can be done = 5 $\times$ 4 = 20

Answer: Option D. -> m!m+1Pn        
:
D
The number of ways in which they can be seated = $m!{.}^{m+1}{\text{P}}_n$

Answer: Option B. -> 3!6!
:
B
Consonants occupy 2 ends in ${}^3{\text{P}}_2$ways remaining 6 letters occupy 6 places in 6! Ways
So the required number of arrangements = ${}^3{\text{P}}_2.6!=3!6!$

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.
$\therefore$The total number of such 5 digit number = ${}^5{P}_5+({}^5{P}_5-{}^4{P}_4)$= 120 + 96 = 216 .