In the word APPLE, there are 5 letters
The lexicographic order of the word is A, E, L, P, P
The letter P occurs 2 times.
Number of words starting with AE = \(\frac{3!}{2}=3\)
Number of words starting with AL  =\(\frac{3!}{2}=3\)
Number of words starting with APE =2! = 2
Number of words starting with APL = 2! = 2
APPEL =1
APPLE =1
Rank of APPLE = 3 + 3 + 2 + 2 + 1 + 1 = 12
Hence proved.

Let p(n) : 2n < (n + 2)! for all k\(\in\)N.
Step1: P(1) 2.1 < (1+2)!
\(\Rightarrow\) 2 < 3!\(\Rightarrow\) 2 < 6 which is true for P(1)  [\(\because\) 3! = 3 x 2 x 1 = 6]
Step 2: P(k) : 2k < (k + 2)!. Let it be true for P(k)
Step3:P(k+1) 2(k+1) < (k+1+2)!
Since 2k < (k + 2)!  (from Step 2)
\(\Rightarrow\)2k+2 < (k+2)! +2
\(\Rightarrow\) 2(k+1) < (k+2)!+2
Also, (k+ 2)! + 2 < (k+ 3)!
\(\therefore\) 2(k+1) < (k+3)!
\(\Rightarrow\) 2(k + 1) < (k + 2 + 1)! which is true for P(k + 1)
Hence, P(k + 1) is true whenever P(k) is true.

Here ²ⁿC₃ : ⁿC₂ = 12 : 1
\(\Rightarrow {(2n)!\over 3!(2n-3)!}\times {2!(n-2)!\over n!}={12\over 1}\)
\(\Rightarrow {(2n)(2n-1)(2n-3)!\over 3\times 2!(2n-3)!}\times {2!(n-2)!\over n(n-1)(n-2)!}={12\over 1}\)
\(\Rightarrow {(2n)(2n-1)(2n-2)\over 3 }\times {1\over n(n-1)}={12\over 1}\)
\(\Rightarrow {4(2n-1)\over 3}={12\over1}\)
\(\Rightarrow \) 8n - 4 = 36\(\Rightarrow \)n = 5

There are 12 letters in the given word of which 2 are T's and the remaining are distinct.
Remaining 10 letters between P and S can be arranged in \(\frac{10!}{2!}\) ways.
ஃ Total number of words starting with P and ending is S = \(\frac{10!}{2!}\) 
= \(\frac { 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2! }{ 2! } \)
= 1814400

Clearly a number between 100 and 1000 has 3 digits.
\(\therefore\) Total number of 3 digit numbers having atleast one of their digits as y.
= (Total number of 3 digit numbers) - (Total number of 3 digit numbers in which y does not appear at all)
Total number of 3 digit numbers' we have to form 3 digit numbers by using the digits 0, 1, 2, 3, ... 9. Hundreds' place can be filled in 9 ways and each of the ten's and one's place can be filled in 10 ways. So, total number of 3 digit numbers = 9 \(\times\) 10 \(\times\) 10 = 900.
Total number of 3 digit number in which 7 does not appear. Here, we have to form 3-digit numbers by using the digits 0 to 9, except 7. So, hundreds place can be filled in 8 ways, and each of the ten's and one's place can be filled in 9 ways.
So, total number of 3-digit numbers in which 7 does not appear at all is 8 \(\times\) 9 \(\times\) 9.
Hence, total number of 3-digit numbers having atleast one of their digits as 7 is 9 \(\times\) 10 \(\times\) 10 - 8 \(\times\)9 \(\times\)9
= 900 - 648 = 252