7 (a) Using $nC_r = \frac{n!}{r!(n-r)!}$ show that $nC_2 = \frac{n(n-1)}{2}$ [2 marks] 7 (b) (i) Show that the equation 2 $\times C_4 = 51 \times C_2$ simplifies to $n^2 - 5n - 300 = 0$ [3 marks] 7 (b) (ii) Hence, solve the equation 2 $\times C_4 = 51 \times C_2$ [2 marks] - AQA - A-Level Maths Mechanics - Question 7 - 2020 - Paper 3

We start with the equation:

[ 2 \times C_4 = 51 \times C_2 ]

Expressing both combinations using the formula ( nC_r = \frac{n!}{r!(n-r)!} ):

[ C_4 = \frac{n!}{4!(n-4)!} \quad \text{and} \quad C_2 = \frac{n!}{2!(n-2)!} ]

So we get:

[ 2 \times \frac{n!}{4!(n-4)!} = 51 \times \frac{n!}{2!(n-2)!} ]

Cancelling ( n! ) from both sides:

[ 2 \times \frac{1}{4!(n-4)!} = 51 \times \frac{1}{2!(n-2)!} ]

Multiplying both sides by ( 4!(n-4)! ) gives:

[ 2 = 51 \times \frac{4!}{2!} \times \frac{1}{(n-2)(n-3)} ]

Since ( 4! = 24 ) and ( 2! = 2 ):

[ 2 = 51 \times 12 \times \frac{1}{(n-2)(n-3)} ]

This simplifies to:

[ (n-2)(n-3) = 306 ]

Expanding gives:

[ n^2 - 5n - 300 = 0 ]

To solve the quadratic equation:

[ n^2 - 5n - 300 = 0 ]

We can use the quadratic formula:

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

In this case, ( a = 1, b = -5, c = -300 ):

[ b^2 - 4ac = (-5)^2 - 4 \times 1 \times (-300) = 25 + 1200 = 1225 ]

So,

[ n = \frac{5 \pm \sqrt{1225}}{2} ]

Calculating the square root of 1225 is:

[ \sqrt{1225} = 35 ]

Thus,

[ n = \frac{5 \pm 35}{2} ]

This gives two possible solutions:

[ n = \frac{40}{2} = 20
\text{or}
n = \frac{-30}{2} = -15 ]

Since ( n ) must be a positive integer, we have ( n = 20 ) as the solution.