I have been stuck in this working out for 2 hours. Can anyone explain ( working out would be great ) how did he get 2(n-2)! = 4 * 6(n-3)! ?
View attachment 10634
Well, I am lost.
We start with a summation symbol referencing k, but nothing to sum and neither a starting nor ending value for k. We have a coefficient of zero to the n? Huh? Is this some sort of notation that I am unfamiliar with?
In any case, how do we possibly "know" from the binomial theorem that
\(\displaystyle \dbinom{n}{3} = 4 * \dbinom{n}{3}.\)
\(\displaystyle \dbinom{n}{3} = 4 * \dbinom{n}{3} \implies 4 = 1 \text { or } \dbinom{n}{3} = 0.\)
Nor is it generally true that
\(\displaystyle \dfrac{n!}{3! * (n - 3)!} = 4 * \dfrac{n!}{2! * (n - 2)!}\)
For example,
\(\displaystyle \dfrac{6!}{3! * 3!} = \dfrac{6 * 5 * 4}{3 * 2} = 20.\) But
\(\displaystyle 4 * \dfrac{6!}{2! * 4!} = 4 * \dfrac{6 * 5}{2} = 4 * 15 = 60.\)
I agree of course that
\(\displaystyle \dbinom{14}{3} = \dfrac{14!}{3! * 11!} = \dfrac{14 * 13 * 12}{3 * 2} =4 * \dfrac{14 * 13 * 3}{3 * 2} =\)
\(\displaystyle 4 * \dfrac{14 * 13}{2} = 4 * \dfrac{14 * 13 * 12!}{2 * 12!} = 4 \dfrac{14!}{2! * 12!} = 4 * \dbinom{14}{2}.\)
Bizarre.