$\mathrm{1.3.5......}(2n-1)2^n$ = $\frac{\mathrm{1.2.3.4.5.6....}(2n-1)\left(2n\right)2^n}{\mathrm{2.4.5.....2}n}$

= $\frac{\left(2n\right)!2^n}{2^n\left(\mathrm{1.2.3......}n\right)}$ = $\frac{\left(2n\right)!}{n!}$

To find the number of times 3 occurs in listing the integer from 1 to 999. (since 3 does not occur in 1000). Any number between 1 to 999 is a 3 digit number xyzz where the digit x,y,z are any digits from 0 to 9.

Now, we first count the numbers in which 3 occurs once only. Since 3 can occur at one place in ${}^3{C}_1$ ways, there are ${}^3{C}_1.(9\times 9)$ = ${3.9}^2$ such numbers.

Again, 3 can occur in exactly two places in ${}^3{C}_1\left(9\right)$ such numbers. Lastly 3 can occur in all the three digits in one such number only 333.

$\therefore$The number of times 3 occurs is equal to $1\times (3\times 9^2)+2\times (3\times 9)+3\times 1$=300. 

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 .

The total number of 4 digits are 9999-999=9000.

The numbers of 4 digits number divisible by 5 are $90\times 20$=1800. Hence required number of ways are 9000-1800 =7200.

{Since there are 20 numbers in each hundred (1 to 100) divisible by 5 and from 999 to 9999 there are 90 hundreds, hence the results}.

Since at any place, any of the digits 2, 5 and 7 can be used, total number of such positive n-digit numbers are $3^n$. Since we have to form 900 distinct numbers, hence $3^n\le 900\Rightarrow n$ = 7

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.

$\circ T\circ R\circ N\circ G\circ L\circ$

Three vowels can be arrange at 6 places in ${}^6{P}_3$ = 120 ways. Hence the required number of arrangements = $120\times 5!$ =14400.

$\frac{56!}{(50-r)!}\times \frac{(51-r)!}{54!}$

= $\frac{30800}{1}\Rightarrow 56\times 55\times (51-r)$ = $30800\Rightarrow r$ = 41

The word ARRANGE, has AA,RR, NGE letters. That is two A' s, two R's and N, G, E one each.
$\therefore$The total number of arrangements $\frac{7!}{2!2!1!1!1!}$=1260 
But, the number of arrangements in which both RR are together as one unit =  $\frac{6!}{2!1!1!1!1!}$ = 360
$\therefore$The number of arrangements in which both RR do not come together = 1260 - 360 = 900. 

Since no two lines are parallel and no three are concurrent, therefore n straight lines intersect at ${}^n{C}_2$ = N (say) points. Since two points are required to determine a straight line, therefore the total number of lines obtained by joining N points ${}^N{C}_2$. But in this each old line has been counted ${}^{n-1}{C}_2$ times, since on each old line there will be n-1 points of intersection made by the remaining (n-1) lines.

Hence the required number of fresh lines is ${}^N{C}_2-n.{}^{n-1}{C}_2$ = $\frac{N(N-1)}{2}-\frac{n(n-1)(n-2)}{2}$

= $\frac{{}^n{C}_2({}^n{C}_2-1)}{2}-\frac{n(n-1)(n-2)}{2}$ = $\frac{n(n-1)(n-2)(n-3)}{8}$