1. ## Combinations (NcR)

is this true for all k?

k(n on top of r) = (kn on top of kr)
if yes, explain why, if no, isit true for any k?

NB: when i say 'on top of' i mean this kind of 'on top of' Combination - Wikipedia, the free encyclopedia
the big bracket with the floating letters inside with no lines

my first answer was k can be anything but 0, but then because 0! is 1 so my answer now is k can be any number, including neg numbers, what are your thoughts? thanks.

2. Originally Posted by LeiaBrown
is this true for all k?
k(n on top of r) = (kn on top of kr)
if yes, explain why, if no, isit true for any k?
To be clear without wasting time.
Are we assuming that $k,~n,~\&~r$ are positive integers?

And the question is $k\binom{n}{r}=\binom{kn}{kr}~?$

BTW: $$k\binom{n}{r}=\binom{kn}{kr}~?$$ gives $k\binom{n}{r}=\binom{kn}{kr}~?$

3. yes thats the question ^

sorry i forgot "r=2"
k, n and r are positive integers
(and thanks for the [tex] thing)

4. Originally Posted by LeiaBrown
yes thats the question ^
sorry i forgot "r=2"
k, n and r are positive integers
Well it does not work: $6\dbinom{10}{2}=240$ and $\dbinom{60}{12}=1399358844975$

5. Originally Posted by LeiaBrown
yes thats the question ^

sorry i forgot "r=2"
k, n and r are positive integers
(and thanks for the [tex] thing)
Is it true that $\displaystyle 2 {3 \choose 1} = {6 \choose 2}$ ?

6. ok thanks, im still a bit confused but thanks anyway

7. Originally Posted by LeiaBrown
ok thanks, im still a bit confused but thanks anyway
What is there to be confused about? You've been given two examples (known as counter examples) of it not working .... Draw a conclusion.

$\dbinom{10}{2}=\dfrac{10\cdot 9}{2\cdot 1}=45$
$\dbinom{20}{4}=\dfrac{20\cdot 19\cdot 18\cdot 17}{4 \cdot 3\cdot 2\cdot 1}=5\cdot 19 \cdot 3\cdot 17$.