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

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 C. -> 10.27
:
C
The number of selections = coefficient of $x^8$ in
$(1+x+x^2+..........+x^8)(1+x+x^2+.........+x^8).(1+x{)}^8$
= coefficient of $x^8$ in $\frac{(1-x^9{)}^2}{(1-x{)}^2}(1+x{)}^8$
= coefficient of $x^8$ in $(1+x{)}^3(1-x{)}^{-2}$
= coefficient of $x^8$ in
$({}^8{C}_0+{}^8{C}_{1}x+{}^8{C}_2{x}^2+..........+{}^8{C}_8{x}^8)\times (1+2x+3x^2+4x^3+.........+9x^8+.........)$
= $9.{}^8{C}_0+8.{}^8{C}_1+7.{}^8{C}_2+.............+1.{}^8{C}_8$
= $C_0+2C_{1}x+{}^3{C}_2{x}^2+......{}^9{C}_8{x}^8$ = $(1+x{)}^8+8x(1+x{)}^7$
Putting x = 1, we get $C_0+2C_1+3C_2+........+9C_8$
= $2^8+{8.2}^7$ = $2^7.(1+8)$ = ${10.2}^7$.

Answer: Option B. -> 35
:
B
Required number of ways
= coefficient of $x^{16}$ in $(x^3+x^4+x^5+.......x^7{)}^4$
= coefficient of $x^{16}$ in $x^{12}(1+x+x^2+.......x^4{)}^4$
= coefficient of $x^{16}$ in $x^{12}(1-x^5{)}^4(1-x{)}^{-4}$
= coefficient of $x^4$ in $(1-x^5{)}^4(1-x{)}^{-4}$
= coefficient of $x^4$ in $(1-4x^5+.....)$
$[1+4x+....+\frac{(r+1)(r+2)(r+3)}{3!}x]$
= $\frac{(4+1)(4+2)(4+3)}{3!}$ = 35
Aliter: Remaining 4 rupees can be distributed in ${}^{4+4-1}{C}_{4-1}$ i.e., 35 ways

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. -> n(n−1)(n−2)(n−3)8
:
C
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 ${}^N{C}_2-n.{}^{n-1}{C}_2$ = $\frac{N(N-1)}{2}-\frac{n(n-1)(n-2)}{2}$
= $\frac{{}^n{C}_2({}^n{C}_2-1)}{2}-\frac{n(n-1)(n-2)}{2}$ = $\frac{n(n-1)(n-2)(n-3)}{8}$

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

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. -> 900
:
B
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 bothRR 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.