Question

prove that (n choose k):(n choose k+1) = (k+1):(n-k)

Answer 1

Do you mean
\[\frac{\dbinom nk}{\dbinom n{k+1}}=\frac{k+1}{n-k}~~?\]
so that the colons (:) represent ratios?

Answer 2

yes

Answer 3

Alright, let's see what you get if you use the binomial coefficient formula:
\[\begin{align*}\frac{\dbinom nk}{\dbinom n{k+1}}&=\frac{\dfrac{n!}{k!(n-k)!}}{\dfrac{n!}{(k+1)!(n-(k+1))!}}&\text{replacement with formula}\\\\
&=\frac{\dfrac{1}{k!(n-k)!}}{\dfrac{1}{(k+1)!(n-(k+1))!}}\\\\
&=\frac{(k+1)!(n-(k+1))!}{k!(n-k)!}\\\\
&=\frac{(k+1)k!(n-(k+1))(n-k)!}{k!(n-k)!}&\text{factor out the first term}\\\\
&=(k+1)(n-(k+1))\end{align*}\]
Hmm... are you sure about that answer?

Answer 4

Just checked, the answer you're supposed to get is right. I must have made some slight mistake, let me look this over...

Answer 5

yes @SithsAndGiggles , i found the mistake.

Answer 6

Oh, got it, I pulled out the wrong factor:
\[\begin{align*}\frac{\dbinom nk}{\dbinom n{k+1}}&=\frac{\dfrac{n!}{k!(n-k)!}}{\dfrac{n!}{(k+1)!(n-(k+1))!}}\\\\
&=\frac{\dfrac{1}{k!(n-k)!}}{\dfrac{1}{(k+1)!(n-(k+1))!}}\\\\
&=\frac{(k+1)!(n-(k+1))!}{k!(n-k)!}\\\\
&=\frac{(k+1)k!(n-(k+1))!}{k!(n-k)!}\\\\
&=\frac{(k+1)\color{red}{((n-k)-1)!}}{(n-k)!}\\\\
&=\frac{(k+1)((n-k)-1)!}{(n-k)((n-k)-1)!}\\\\
&=\frac{k+1}{n-k}\end{align*}\]

Answer 7

yes,that's the mistake ;)

Answer 8

and @pulliamk , u can prove this by using combinatorics too
@SithsAndGiggles has proved it by using algebra....but the combinatorics proof is very difficult and this is easier ;)

Answer 9

@PFEH.1999 yes I considered trying the combinatorial proof but I figured the algebraic method is easier to follow. Not to mention it's just plain hard :P

Answer 10

but i think algebra cannot handle more difficult problems ;)

Answer 11

Oops ! i deleted it :(

Answer 12

help me

Answer 13

my exam is very near so I need you help

Answer 14

Garde 12 mathematics. Do you know how to do calculas part I am having trouble