Any number greater than a million will contain all the seven digits.
Now, we have to arrange these seven digits, out of which 2 occur twice, 3 occurs twice and the rest are distinct.
The number of such arrangements =\(\frac { 7! }{ 2!3! } =\frac { 7\times 6\times 5\times 4\times 3! }{ 2\times 3! } =420\)
These arrangements also include those numbers which contain 0 at the million's place.
Keeping 0 fixed at the million place, we have 6 digits out of which 2 occurs twice, 3 occurs thrice and the rest are distinct.
These 6 digits can be arranged in \(\frac { 6! }{ 2!3! } =\frac { 6\times 5\times 4\times 3! }{ 2\times 1\times 3! } =60\quad ways\)
Hence, the number of required numbers = 420 - 60 = 360.

Given 22 P_{r + 1} :20 P_{r+ 2} = 11 : 52
\(\Rightarrow\)52.22P_r+1 = 11.20 P_r+2 
\(\Rightarrow\)    \(52.\frac { 22! }{ (22-r-1)! } =11.\frac { 20! }{ (20-r)-2! } \quad \left[ \therefore npr=\frac { n! }{ n-r! } \right] \)
\(\Rightarrow\)    \(\frac { 52\times 2\times 21 }{ (21-r)! } =11.\frac { 20! }{ (18-r)! } \)
\(\Rightarrow\)    \(\frac { 52\times 2\times 21 }{ (21-r)(20-r)(19-r)(18-r)! } =\frac { 1 }{ (18-)! } \)
\(\Rightarrow\)    52 x 2 x 21 = (21- r) (20 - r) (19- r)
\(\Rightarrow\)    2 x 3 x 7 x 4 x 13 = (21- r) (20 - r) (19- r)
\(\Rightarrow\)    12 x 13 x 14 = (21- r)(20 - r)(19- r)
\(\Rightarrow\)    12 x 13 x 14 = (19- r) (20 - r) (21- r)
\(\Rightarrow\)    19-r=12
\(\Rightarrow\)r= 19-12
∴  r =7

A number between 100 and 1000 has 3 digits.
∴  Total number of 3 digit numbers having at least one of their digits as 7.
= (Total no. of 3 digit numbers) -(Total no.of 3 - digit numbers in which 7 does not appear at all)
Total no.of  3 - digit numbers : 

The given digits are 0, 1,2,3,4,5,6, 7, 8, 9
Hundred's place can be filled in 9 ways (excluding 0) and tens and ones place can be filled in 10 ways.
∴  Total number of 3 digit numbers = 9 x 10 x 10 = 900
Total no. of 3 - digit numbers in which 7 does not appear at all : 

The given digits are 0, 1,2,3,4,5,6, 7, 8, 9
Hundred's place can be filled in 8 ways, tens and ones place can be filled in 9 ways.
∴  Total number of 3 digit numbers in which 7 does not appear at all = 8 x9 x 9 = 648
∴  Required numbers = 900 - 648 = 252

I and II prizes Mathematics can be given in (30 x 29) ways.
I & II prizes in physics also can be given in (30 x 29) ways.
I prize in chemistry can be given in 30 ways.
I prize in English can be given in 30 ways.
Hence, the number of ways to give prizes in all the four subjects = (30 x 29) x (30 x29)x 30 x 30
= 6.8121 x 10⁸

Since the numerator degree is equal to the degree of the denominator, let us divide.

\(\therefore\)   \({{x^2+x+1}\over{x^2+2x+1}}=1-{{x}\over{x^2+2x+1}}\)  ...(1)
Consider    \({{x}\over{x^2+2x+1}}={{x}\over{(x+1)^2}}={{A}\over{x+1}}+{{B}\over{{(x+1)}^{2}}}\)
\(\therefore\ {{x}\over{x^2+2x+11}}={{A(x+1)+B}\over{{(x+1)}^{2}}}\)
\(\Rightarrow\)      x = A(x+ 1) + B  ..(2)
Putting x = - 1 in (2) we get
-1 = 0 + B   \(\Rightarrow\ \ \boxed{B=-1}\)
Putting x = 0 in (2) we get,
0 =A+B \(\Rightarrow\) A - 1 = 0  \(\Rightarrow\)  \(\boxed{A=1}\)
\(\therefore\)   \({{x}\over{{(x+1)}^{2}}}={{1}\over{x+1}}-{{1}\over{{(x+1)}^{2}}}\) ..(2)
Substituting (2) in (1) we get,
\({{x^2+x+1}\over{x^2+2x+1}}=1-{{1}\over{x+1}}+{{1}\over{{(x+1)}^{2}}}\)