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Prove: \[\binom{n-1}{k-1} +\binom{n-2}{k-1} +\binom{n-3}{k-1} +....+\binom{k-1}{k-1} =\binom{n}{k}\]

Mathematics
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i will guess induction
induction? is there by chance another method?
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"

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Other answers:

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
for base case it is probably easiest to start at \(n=2\)
can it be a gemoetric progression?
geometric*
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
there is a think a way to do it by thinking, however it is almost like saying \[\dbinom{n}{k}=\dbinom{n-1}{k}+\dbinom{n-1}{k-1}\]
hm..so induction is the only method? but i havent learnt induction to prove these kind of things..
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.
well, thank you!

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