Answer: Option A. -> 7200
:
A
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}.

Answer: Option C. -> 14400
:
C
$\circ T\circ R\circ N\circ G\circ L\circ$
Three vowels can be arrange at 6 places in ${}^6{P}_3$ = 120 ways. Hence the required number of arrangements = $120\times 5!$ =14400.

Answer: Option B. -> 300
:
B
To find the number of times 3 occurs in listing the integer from 1 to 999. (since 3 does notoccur 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.

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. -> 7
:
B
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

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

Answer: Option A. -> 485
:
A
Given, X has 7 friends, 4 of them are ladies and 3 are men while Y has 7 friends, 3 of them are ladies and 4 are men.
$\therefore$ Total number of required ways
${}^3{C}_3{\times }^4{C}_0{\times }^4{C}_0{\times }^3{C}_3{+}^3{C}_2{\times }^4{C}_1{\times }^4{C}_1{\times }^3{C}_2{+}^3{C}_1{\times }^4{C}_2{\times }^4{C}_2{\times }^3{C}_1{+}^3{C}_0{\times }^4{C}_3{\times }^4{C}_3{\times }^3{C}_0=1+144+324+16=485$

Answer: Option A. -> 185
:
A
Required number of ways = ${}^{12}{C}_3-{}^7{C}_3$
= 220 - 35 = 185

Answer: Option A. -> 106
:
A
Required number of lines
= ${}^{16}{C}_2-{}^6{C}_2$ +1 = 120 -15+1= 106