(i) a particular teacher is included?
There are 5 teachers and 20 students 2 teachers out of 5 teachers can be selected in 5C₂ ways.
3 students out of 20 students can be selected in 20C₃ ways
Hence, total number of committees = 20C₃ \(\times \) 5C₂
= \(\frac { 20\times 19\times 18 }{ 3\times 2\times 1 } \times \frac { 5\times 4 }{ 2\times 1 } \)
= \(10\times 19\times 6\times 5\times 2\)
= 11400
Since a particular teacher is included, the committee will have 1 teacher and 3 students.
∴ 1 teacher can be selected from 4'teachers in 4C₁ = 4 ways.
3 students out of 20 students can be selected in 20C₃ ways.
Hence, required number of committees
= 4C₁ \(\times \) 20C₃
= \(4\times \frac { 20\times 19\times 18 }{ 3\times 2\times 1 } \)
= 4560
(ii) 2 teachers can be selected from 5 teachers in 5C₂ ways
Since a particular student is excluded, 3 students can be selected from 19 students in 19C₃ ways
Hence required number of committees = 19C₃ \(\times \) 5C₂
=\(\frac { 19\times 18\times 17 }{ 3\times 2\times 1 } \times \frac { 5\times 4 }{ 2\times 1 } \)
= 19 \(\times \)6\(\times \)17\(\times \)5
= 9690

⁽ⁿ⁺²⁾P₄ = 42 \(\times\) ⁿP₂ 
⇒ \(\frac { ^{ (n+2) }{ P }_{ 4 } }{ ^{ n }P_{ 2 } } \) = 42
⇒ \(\frac { (n+2)(n+1)(n)(n-1) }{ n(n-1) } \) = 42
⇒ (n + 2) (n + 1) = 42 = 7 \(\times\) 6
⇒ n + 2 = 7 
⇒ n = 5

The given strings contains 6 letters (V, O, W, E, L, S).
(i) Since all strings must begin with E, we have the remaining 5 letters which can be arranged in ⁵P₅ = 5! ways.
Therefore the total number of strings with E as the starting letter is 5! =120.

(ii) Since all strings must begin with E, and end with W, we need to fix E and W. The remaining 4 letters can be arranged in ⁴P₄ = 4! Ways.

Therefore the total number of strings with E as the starting letter and W as the final letter is 4! = 24.

Let us arrange the 8-non-mathematics candidates in ⁸P₈ = 8! ways. Each of these arrangements create 9 gaps.
Therefore, the 7 mathematics candidates can be placed in these 9 gaps in ⁹P₇ ways.
_O₁_O₂_O₃_O₄_O₅_O₆_O₇_O_{8_}
By the rule of product, the required number of arrangements is 8! \(\times\) ⁹P₇
= 8! \(\times\) \(\frac { 9! }{ 2! } =\frac { 8!\times 9! }{ 2! } \).

There are 6 places in that there are 3 even places we have 2, 4, 2 as even numbers. The number of ways of permuting 2, 4, 2 in the 3 even places in \(\frac { 3! }{ 2! } \) = 3 ways. The remaining numbers 1, 3, 1 can be permuted in the remaining 3 places in \(\frac { 3! }{ 2! } \) = 3 ways. Hence, the required number of numbers is 3 x 3 = 9.

¹⁰C₂ + 2 x ¹⁰C₃ + ¹⁰C₄ = ¹⁰C₂ + (¹⁰C₃ + ¹⁰C₃) + ¹⁰C₄
= (¹⁰C₂ + ¹⁰C₃) + (¹⁰C₃ + ¹⁰C₄)
= ¹¹C₃ + ¹¹C₄
= ¹²C₄.

The picking of 5 brands for a first, second, third, fourth and fifth best brand from 20 brands in ²⁰P₅ ways. From the remaining 15 we need to select 7 acceptable in ¹⁵C₇ ways. By the rule of product it can be done in ²⁰P₅ x ¹⁵C₇ ways.

Whenever we select 2 lines from the first set of m lines and 2 lines from the second set of n lines, one parallelogram is formed. Thus the number of parallelograms formed is ^mC₂ \(\times\) ⁿC₂.

A polygon of n sides has n vertices. By joining any two vertices of a polygon, we obtain either a side or a diagonal of the polygon: Number of line segments obtained by joining the vertices of a n sided polygon taken two at a time is  ⁿC₂ = \(\frac { n(n-1) }{ 2 } \). Out of these lines, there are n sides of polygon. Therefore, number of diagonals of the polygon is \(\frac { n(n-1) }{ 2 } -n=\frac { n(n-3) }{ 2 } .\)
In particular, for a pentagon and heptagon (Septagon), number of diagonals respectively are \(\frac { 5(5-3) }{ 2 } \) = 5 and \(\frac { 7(7-3) }{ 2 } =14.\)

Let p(n) be the statement that 3n² > (n + 1)² with n\(\ge\) 2. Therefore the first stage is n = 2.
Now, P(2) = 3 x 2² = 12 and 3² = 9. As 12> 9 we get P(2) is true.
We assume that p(n) is true for n = k.
Now,
P(k+ 1) 3(k+ 1)² = 3k² + 6k+ 3
P(k) + 6k+ 3
> (k + 1)² + 6k + 3
k² + 8k+4
k² + 4k+4 + 4k
(k+ 2)² + 4k
> (k+ 2)² since k> 0.
This is the statement P(k + 1). The validity of P(k + 1) follows from that of P(k).
Therefore by the principle of mathematical induction, for all n\(\ge\) 2, 3n² > (n + 1)².