7 (a) Using $C_r^n = \frac{n!}{r!(n - r)!}$ show that $C_2^n = \frac{n(n - 1)}{2}$ - AQA - A-Level Maths Pure - Question 7 - 2020 - Paper 3

To simplify the equation, we first express both combinations:

$C_4^n = \frac{n!}{4!(n - 4)!}$

Substituting $4! = 24$, we have:

$C_4^n = \frac{n!}{24(n - 4)!}$

Thus, substituting in the original equation gives:

$2 \times \frac{n!}{24(n - 4)!} = 51 \times \frac{n(n - 1)}{2}$

This simplifies to:

$\frac{n!}{12(n - 4)!} = \frac{51n(n - 1)}{2}$

Multiplying both sides by $12(n - 4)!$:

$n! = 306n(n - 1)(n - 4)!$

Now, we can express $n!$ as:

$n! = n(n - 1)(n - 2)(n - 3)(n - 4)!$

Thus:

$n(n - 1)(n - 2)(n - 3) = 306$

After simplifying the resulting polynomial, we find:

$n^2 - 5n - 300 = 0$

Now, we can solve the equation:

$n^2 - 5n - 300 = 0$

We use the quadratic formula:

$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1, b = -5, c = -300$. Plugging these values in gives:

$n = \frac{5 \pm \sqrt{(-5)^2 - 4 \times 1 \times (-300)}}{2 \times 1}$

$= \frac{5 \pm \sqrt{25 + 1200}}{2}$

$= \frac{5 \pm \sqrt{1225}}{2}$

Since $$ must be a positive integer, we consider:

$n = \frac{5 + 35}{2} = 20$

Thus, the solution is:

$n = 20$