P(n): "n² + n is even"
Given that P(k) is true.
\(\Rightarrow \)   k² + k is even
\(\Rightarrow \)  k² + k = 2\(\lambda \) for some \(\lambda \)  ....(1)
To prove that P (k + 1) is true.
P (k + 1) : (k + 1)²+ (k + 1) is even
Consider (k + 1)² + (k + 1)
= k²+2k+ 1 +k+ 1
= k²+2k+ 1 +k+ 1
= (k²+ k) + (2 k + 2)
= 2\(\lambda \) + 2 (k + 1)   [from (1)]
= even + even
= Sum of two even numbers is always an even number .
∴  P (k + 1) is true.

P(n): "12n + 3" is a multiple of 5.
P(3): 12 (3) + 3 36 + 3 = 39 is not a multiple of 5.
∴ P (3) is false
P(6) = 12(6) + 3 = 72 + 3 = 75 = 5 x 15 is a multiple of 5.
∴ P (6) is true

Here 12 things are to be divided equally among 4 persons, hence the groups are regarded as distinct.
Hence required number of ways = \(\frac { 12! }{ { (3!) }^{ 4 } } \)
\(=\frac { 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 }{ 3\times 2\times 3\times 2\times 3\times 2\times 3\times 2 } \)
=3696000

No. of ways of selecting 4 questions out of 6 questions (Part - I) = 6C₄ = 6C₂ =\(\frac { 6\times 5 }{ 1\times 2 } =15\)
No. of ways of selecting 4 questions out of 7 questions (Part - II) = 7C₄ = 7C₃ =\(\frac { 7\times 6\times 5 }{ 3\times 2\times 1 } =35\)
No. of ways of selecting 4 questions out of 8 questions (Part - III) = 8 C₄ =\(\frac { 8\times 7\times 6\times 5 }{ 4\times 3\times 2\times 1 } =70\)
∴ Total number of choices = 15 x 35 x 70 = 36750

No. of ways of giving I prize = n
No. of ways of giving II prize = n
.............................................
Similarly no. of ways of giving n^th prize = n
Therefore By fundamental principle of counting, the required number of ways