Aha I understand now, I need to use that formula for those types of questions.
I do understand how to count with factorials, but what do I do if there are undefined numbers n! & n. Do I need to make up numbers for n until I get an positive integer?
How do you find unknowns? You use algebra. You have one unknown, n, and one equation. All is good.
[MATH]\dbinom{n}{2} = \dbinom{n}{3} \implies \dfrac{n!}{2! * (n - 2)!} = \dfrac{n!}{3! * (n - 3)!} \implies[/MATH]
[MATH]\dfrac{n!}{2 * (n - 2)!} = \dfrac{n!}{6 * (n - 3)!}.[/MATH]
This does not even make sense unless n > 2.
Now you need to know an
IMPORTANT theorem.
[MATH]\dfrac{n!}{(n - k)!} = \prod_{j=0}^{k - 1} (n - j) \text { if } n > k > 0.[/MATH]
That great big pi means to multiply a series of numbers. For example,
[MATH]\prod_{j=1}^3 x_j = x_1 * x_2 * x_3.[/MATH]
So let's put that to work in our problem.
[MATH]\dfrac{n!}{2 * (n - 2)!} = \dfrac{n!}{6 * (n - 3)!} \implies \dfrac{\displaystyle \prod_{j=0}^{2 - 1} (n - j)}{2} = \dfrac{\displaystyle \prod_{j=0}^{3 - 1} (n - j)}{6} \implies[/MATH]
[MATH]\dfrac{\displaystyle \prod_{j=0}^{1} (n - j)}{2} = \dfrac{\displaystyle \prod_{j=0}^{2} (n - j)}{6} =\dfrac{(n - 0) * (n - 1)}{2} = \dfrac{(n - 0) * (n - 1) * (n - 2)}{6} \implies[/MATH]
[MATH]6n(n - 1) = 2n(n - 1)(n - 2) \implies 6 = 2(n - 2) \implies n = 5.[/MATH]
LETS CHECK.
[MATH]\dbinom{5}{2} = \dfrac{5!}{2! * (5 - 2)!} = \dfrac{5 * 4 * \cancel {3!}}{2 * \cancel {3!}} = \dfrac{5 * 4}{2} = 10.[/MATH]
[MATH]\dbinom{5}{3} = \dfrac{5!}{3! * (5 - 3)!} = \dfrac{5 * 4 * 3 * \cancel {2!}}{6 * \cancel {2!}} = \dfrac{5 * 4 * 3}{6} = 10.[/MATH]
