Evaluate the following:
(i) `.^(10)C_(5)` (ii) `.^(12)C_(8)`
(iii) `.^(15)C_(12)` (iv) `.^(n+1)C_(n)`
(v) `.^(14)C_(9)`

Answer

To evaluate the combinations given in the question, we will use the formula for combinations, which is:

\[ nCr = \frac{n!}{r!(n-r)!} \]

Now, let's solve each part step by step.

### (i) Evaluate \(^{10}C_{5}\)

Using the formula:

\[ ^{10}C_{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5! \cdot 5!} \]

Now, we can simplify \(10!\):

\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5! \]

Substituting this back into the equation:

\[ ^{10}C_{5} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \cdot 5!} \]

The \(5!\) cancels out:

\[ = \frac{10 \times 9 \times 8 \times 7 \times 6}{5!} \]

Calculating \(5! = 120\):

\[ = \frac{10 \times 9 \times 8 \times 7 \times 6}{120} \]

Calculating the numerator:

\[ 10 \times 9 = 90, \quad 90 \times 8 = 720, \quad 720 \times 7 = 5040, \quad 5040 \times 6 = 30240 \]

Now, divide by \(120\):

\[ = \frac{30240}{120} = 252 \]

So, \(^{10}C_{5} = 252\).

### (ii) Evaluate \(^{12}C_{8}\)

Using the formula:

\[ ^{12}C_{8} = \frac{12!}{8!(12-8)!} = \frac{12!}{8! \cdot 4!} \]

Simplifying \(12!\):

\[ 12! = 12 \times 11 \times 10 \times 9 \times 8! \]

Substituting back:

\[ ^{12}C_{8} = \frac{12 \times 11 \times 10 \times 9 \times 8!}{8! \cdot 4!} \]

The \(8!\) cancels out:

\[ = \frac{12 \times 11 \times 10 \times 9}{4!} \]

Calculating \(4! = 24\):

\[ = \frac{12 \times 11 \times 10 \times 9}{24} \]

Calculating the numerator:

\[ 12 \times 11 = 132, \quad 132 \times 10 = 1320, \quad 1320 \times 9 = 11880 \]

Now, divide by \(24\):

\[ = \frac{11880}{24} = 495 \]

So, \(^{12}C_{8} = 495\).

### (iii) Evaluate \(^{15}C_{12}\)

Using the formula:

\[ ^{15}C_{12} = \frac{15!}{12!(15-12)!} = \frac{15!}{12! \cdot 3!} \]

Simplifying \(15!\):

\[ 15! = 15 \times 14 \times 13 \times 12! \]

Substituting back:

\[ ^{15}C_{12} = \frac{15 \times 14 \times 13 \times 12!}{12! \cdot 3!} \]

The \(12!\) cancels out:

\[ = \frac{15 \times 14 \times 13}{3!} \]

Calculating \(3! = 6\):

\[ = \frac{15 \times 14 \times 13}{6} \]

Calculating the numerator:

\[ 15 \times 14 = 210, \quad 210 \times 13 = 2730 \]

Now, divide by \(6\):

\[ = \frac{2730}{6} = 455 \]

So, \(^{15}C_{12} = 455\).

### (iv) Evaluate \(^{n+1}C_{n}\)

Using the formula:

\[ ^{n+1}C_{n} = \frac{(n+1)!}{n!(n+1-n)!} = \frac{(n+1)!}{n! \cdot 1!} \]

This simplifies to:

\[ = \frac{(n+1) \cdot n!}{n!} = n + 1 \]

So, \(^{n+1}C_{n} = n + 1\).

### (v) Evaluate \(^{14}C_{9}\)

Using the formula:

\[ ^{14}C_{9} = \frac{14!}{9!(14-9)!} = \frac{14!}{9! \cdot 5!} \]

Simplifying \(14!\):

\[ 14! = 14 \times 13 \times 12 \times 11 \times 10 \times 9! \]

Substituting back:

\[ ^{14}C_{9} = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9!}{9! \cdot 5!} \]

The \(9!\) cancels out:

\[ = \frac{14 \times 13 \times 12 \times 11 \times 10}{5!} \]

Calculating \(5! = 120\):

\[ = \frac{14 \times 13 \times 12 \times 11 \times 10}{120} \]

Calculating the numerator:

\[ 14 \times 13 = 182, \quad 182 \times 12 = 2184, \quad 2184 \times 11 = 24024, \quad 24024 \times 10 = 240240 \]

Now, divide by \(120\):

\[ = \frac{240240}{120} = 2002 \]

So, \(^{14}C_{9} = 2002\).

### Final Answers:
1. \(^{10}C_{5} = 252\)
2. \(^{12}C_{8} = 495\)
3. \(^{15}C_{12} = 455\)
4. \(^{n+1}C_{n} = n + 1\)
5. \(^{14}C_{9} = 2002\)