Total number of symbols  = 36
There are 36 x 36 x 36 x 36 = (36)⁴ = 1679616 PINs in all.
[by fundamental principle of counting]
Note that, when repetition of symbols is not allowed, then there are  36 x 35 x 34 x 33 = 1413720  different  PINs
Now, the number of PINs that contains at least one repeated symbol = 1679616 - 1413720 = 265896
= 0.1583

We know that, in a chess board, there are 64 squares of which 32 are white and 32 are black.
Clearly, 3 squares can be selected in \(^{ 64 }{ C }_{ 3 }\) ways.
Let E be the event of getting 2 squares of one colour and other of a different colour.
Then, n(E) = \(^{ 32 }{ C }_{ 2 }\times ^{ 32 }{ C }_{ 2 }+^{ 32 }{ C }_{ 1 }\times ^{ 32 }{ C }_{ 2 }\)
[ 2 squares of one colour and 1 of other can be 2w, 1B or 1w , 2B]
\(2\times ^{ 32 }{ C }_{ 2 }\times ^{ 32 }{ C }_{ 1 }\)
\(\frac { 16 }{ 21 } \)

\(\frac { 1 }{ 2 } \)

\(\frac { 1 }{ 2 } \)

\(\frac { 1 }{ 13 } \)

\(f'(x)={ 100x }^{ 99 }+{ 99x }^{ 98 }+...+1+0\)
\(={ 100x }^{ 99 }+{ 99x }^{ 98 }++...+1\)
\(Now,\ f'(x)=100+99+...+1\)
\(=\frac { 100 }{ 2 } \left[ 2\times 100+(100-1)(-1) \right] =5050\)

1:3 externally

Let Since, it is passes through (3,0) and  \((3\sqrt { 2 } ,2)\)
\(\frac { 9 }{ { a }^{ 2 } } -0=1\)  and  \(\frac { 18 }{ { a }^{ 2 } } -\frac { 4 }{ { b }^{ 2 } } =1\)
a² = 9 and b² = 4
b² = a² (e²-1)
\(e=\frac { \sqrt { 13 } }{ 2 } \)

Let the point (h, k) id equidistant from the given lines.
Distance from line (i) = \(\frac { \left| 4h+3k+10 \right| }{ \sqrt { 16+9 } } \)
Distance from linr (ii) = \(\frac { \left| 5h-12k+26 \right| }{ \sqrt { 25+144 } } \)
Distance from linr  (iii) = \(\frac { \left| 7h+24k-50 \right| }{ \sqrt { { 7 }^{ 2 }+{ 24 }^{ 2 } } } \)
Since, the point (h, k) is equidistant from lines (i), (ii) and (iii).
\(\therefore \frac { \left| 4h+3k+10 \right| }{ \sqrt { 16+9 } } =\frac { \left| 5h-12k+26 \right| }{ \sqrt { 25+144 } } =\frac { \left| 7h+24k-50 \right| }{ \sqrt { 49+576 } } \)
\(\Rightarrow \frac { \left| 4h+3k+10 \right| }{ 5 } =\frac { \left| 5h-12k+26 \right| }{ 13 } =\frac { \left| 7h+24k-50 \right| }{ 25 } \)
Clearly, if h=0, k=0, then \(\frac { 10 }{ 5 } =\frac { 26 }{ 13 } =\frac { 50 }{ 25 } =2\)
Ans. (0, 0)

Since, x , 2y and 3z are in AP.
\(\therefore \)  4y = x + 3z
And x, y, z are in GP.
 \(\therefore \)   y = rx and z = xr²
On putting the value of y and z in Eq. (i), we get
4xr = x + 3xr²
\(\Rightarrow \)  3r² - 4r + 1 = 0
r = \(\frac { 1 }{ 3 } \)  [\(\because \) r = 1 is not possible]

Here, r>1, n>2
\(\therefore \)  T_3r = ²ⁿC_3r-1 x^3r-1; T_{r+2 =} ²ⁿC_r+1 x^r+1
Then, ²ⁿC_3r-1= ²ⁿC_r+1 
\(\Rightarrow \) 3r - 1 + r + 1 = 2n \(\Rightarrow \) n = 2r

Total nuber of 4 letters code = ¹⁰C₄

7200

For the equation x+y=20, the shaded area and origin both lies on the same side of the live, therefore the compounding inequality is \(x+y\le 20\).
Similarly, \(3x+2y\le 48\)
Also,the shaded portion lines in Ist quadrant.
\(\therefore \quad x\ge 0\quad and\quad y\ge 0\)
Ans. \(x+y\le 20,3x+2y\le 48,x\ge 0,y\ge 0\)

Let the length of shortest side be x cm
Then, according to question, we have 
Length of third side = (x+2) cm
Since the perimeter of the triangle is more than 166 cm.
2x+x+(x+2)>166
\(\Rightarrow 4x+2>166\)
\( \Rightarrow 4x>164\quad [subtracting\quad 2\quad from\quad both\quad sides]\)
\(\Rightarrow x>41\quad [dividing\quad both\quad sides\quad by\quad 4]\)
Hence,the length of the shortest side should be greater than 41cm.

Let z = x + iy, then
\(\frac { z-1 }{ z+1 } =\frac { ({ x }^{ 2 }-1)+{ y }^{ 2 }+i[y(x+1)-y(x-1)] }{ ({ x }^{ 2 }+1)^{ 2 }+y^{ 2 } } \\ \)
\(\because \frac { z-1 }{ z+1 } \)  is purely imaginary.
\(\therefore Re(\frac { z-1 }{ z+1 } ) =0\ i.e.\ \frac { ({ x }^{ 2 }-1)+{ y }^{ 2 } }{ ({ x }^{ 2 }+1)^{ 2 }+y^{ 2 } } =0\)
\({ x }^{ 2 }-1-{ y }^{ 2 }=0\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }=1= |z|=1\)

\(\frac { 2-i }{ 2 } \)

3i³- 2ai²+(1-a)i + 5
= 3( - i) + 2a + (1 - a)i+5  [i³ = -i and i² = -1]
= (2a + 5) + i(1-a-3), which will be real,
if 1 - a - 3 =0,
i.e, a = - 2

Domain = \((-\infty ,-1)\cup (1,4)\)

Given E = set of students studying English
M =Set of students studying Mathematics
U= Set of all students
Since, all of the students who study Mathematics also study English, but some students who study English do not study Mathematics
\(\therefore \quad \quad M\subset E\subset U\)
Through Venn diagram, we represent it as

The quantifier is 'for all' and the negation is There exists an even integer x such that x² is not even

Direct Method For any real number x,y, it is given
x = y \(\Rightarrow \) 2x = 2y
\(\Rightarrow \) 2x + a = 2y + a for some a \(\in\) Z.
Contrapositive Method The contrapositive statement of 'p' is 'For any real numbers x,y, if 2x+a\(\neq \)2y+a, where a\(\epsilon \)Z.then x\(\neq \)yⁿ .

2k<(k+2)!\(\Rightarrow \)2k+2(k+2)!+2]
\(\Rightarrow \) (k+1)² < 2^k + 2^k    [(2k+1)<2^k for k\(\le \)3]
\(\Rightarrow \) (k+1)² < 2^k+1

Step I: Let P(n) be the given statement
i.e. P(n): \(1\times 1!+2\times 2!+3\times 3!+....+n\times n!=(n+1)!-1\)
Step II : For n=1, we have
LHS=\(1\times 1!\)=1
and RHS=(1+1)!-1=2!-1=2-1=LHS
\(\because \)  LHS=RHS
\(\therefore \)  P(1) is true
Step III  Let us assume that P(n) is true for n=k
Then, we have
P(k): \(1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)!-1\quad \quad ....(i)\)
Step IV  Now, we shall prove the statement for n=k+1. For this we have to show that
\(1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)\times (k+1)!=(k+1+1)!-1\)
Then, LHS \(=1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)\times (k+1)!\)
=(k+1)!-1+(k+1)!\(\times \)(k+1)      [from Eq.(1)]
=(k+1+1)(k+1)!-1=(k+2)(k+1)!-1
=(k+2)!-1   [\(\because \)n(n-1)!=n]
Thus, P(k+1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all natural numbers n.