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## Mimi_x3 3 years ago Prove: $\binom{n-1}{k-1} +\binom{n-2}{k-1} +\binom{n-3}{k-1} +....+\binom{k-1}{k-1} =\binom{n}{k}$

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1. anonymous

i will guess induction

2. Mimi_x3

induction? is there by chance another method?

3. anonymous

maybe a thinking method, but i can't imagine there is another way, since you are making a statement that says essentially "for all $$n\in \mathbb{N}$$ this formula holds"

4. anonymous

the last part will be some annoying algebra with factorials, but since you have a sum with $$n-k$$ terms you cannot add them i in a formula that i can think of. i think the induction step will be enough arithmetic as it is

5. anonymous

for base case it is probably easiest to start at $$n=2$$

6. Mimi_x3

can it be a gemoetric progression?

7. Mimi_x3

geometric*

8. anonymous

maybe, lets try it with numbers , say $$n=5, k=3$$ then $$\dbinom{2}{2}=1,\dbinom{3}{2}=3, \dbinom{4}{2}=6$$ no, not geometric

9. anonymous

there is a think a way to do it by thinking, however it is almost like saying $\dbinom{n}{k}=\dbinom{n-1}{k}+\dbinom{n-1}{k-1}$

10. Mimi_x3

hm..so induction is the only method? but i havent learnt induction to prove these kind of things..

11. joemath314159

You can just keep using:$\left(\begin{matrix}n \\ k\end{matrix}\right)=\left(\begin{matrix}n-1 \\ k\end{matrix}\right)+\left(\begin{matrix}n-1 \\ k-1\end{matrix}\right)$over and over again. In a sense, its an inductive argument. Using it once gives you the above statement. Notice the second part of the sum is something you want, but not the first part. So use that idea on the first one again:$\left(\begin{matrix}n-1 \\ k\end{matrix}\right)=\left(\begin{matrix}n-2 \\ k\end{matrix}\right)+\left(\begin{matrix}n-2 \\ k-1\end{matrix}\right)$So altogether we have:$\left(\begin{matrix}n \\ k\end{matrix}\right)=\left(\begin{matrix}n-2 \\ k\end{matrix}\right)+\left(\begin{matrix}n-2 \\ k-1\end{matrix}\right)+\left(\begin{matrix}n-1 \\ k-1\end{matrix}\right)$Again, the last two parts of the sum are what you want, but the first isnt, so keep repeating this process (this is the inductive reasoning). You do this all the way until the n-2 on the top becomes k-1.

12. Mimi_x3

well, thank you!

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