The definition of $ \binom { newton } { k } $ is not $ \frac { normality ! } { ( n-k ) ! \times k ! } $, that ‘s precisely a formula to calculate it. The conceptual intend of $ \binom { newton } { k } $ is the issue of ways to PICK ( or choose ) $ kelvin $ things out of $ north $, careless of which ordering you pick them in .

To elaborate, let ‘s take an case. Let ‘s say you have $ 5 $ balls called A, B, C, D, and E in a base. You have to select out $ 2 $ of these balls. “ choosing ” $ A, B $ is the like thing as choosing $ B, A $, because the order you pull them out from the bag does not matter. now you can implement the formula, as it is understand why combinations is applied here rather of permutations. Based on the motion you asked and your remark that says “ From what I understand, $ \binom { newton } { k } = \frac { n ! } { ( n-k ) ! \times thousand ! } $ ”, it seems like you have n’t sympathize combinations in depth and memorized the formula to calculate it. I have provided a proof for the recipe in the last section of this post, so that you can understand the conceptual intend of combinations and why the rule works.

so, if we translate your problem in terms of our model, it says that we have $ newton $ balls this fourth dimension called $ 1, 2, 3, \dots, nitrogen $. We have to pick and keep out $ thousand $ of these $ nitrogen $ balls out of the pocket, regardless of which holy order we pick the $ thousand $. We will call the ball we take out of the bag heads, and the remaining balls inside the bag tails. Because of our explanation in the above part, we can similarly use combinations here and not permutations. So it ‘s simply $ \binom { newton } { k } $ .

basically restating my answer to your post, this is good the definition of combinations applied to a problem. No need to use evocation or anything else.

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nowadays, as I said in the second section, I will give the proof of why the formula works.

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Taking our model again, say you have a cup of tea with $ north $ balls in it, and you have to tell the number of ways you can choose a.k.a blue-ribbon a.k.a clean $ k $ balls out of the bag, regardless of which decree you pick the balls in. If the order you pick them would matter, then we would have $ nitrogen ( n-1 ) ( n-2 ) \dots ( n- ( k-1 ) ) = \frac { nitrogen ! } { ( n-k ) ! } $ ways to do it because there are $ north $ ways to pick the first ball, and now there are $ n-1 $ balls left in the bag. then, there are $ n-1 $ ways, and sol on. however, in the veridical problem holy order does NOT topic. In the $ \frac { n ! } { kilobyte ! } $ count, any group of $ kilobyte $ balls have been counted $ k ! $ times as there are $ kelvin ! $ ways to decide which order we will pick some given $ kilobyte $ balls in, and we have counted all of those $ k ! $ ways. however, we only want to count all these $ thousand ! $ consequence as $ 1 $ result, since orderliness does n’t matter in combinations. So we have to divide the original count by $ potassium ! $, and we get $ \frac { \frac { nitrogen ! } { ( n-k ) ! } } { kelvin ! } = \boxed { \frac { north ! } { ( n-k ) ! \times k ! } } $

$ \blacksquare $

**Hopefully, after reading my entire post, you have gotten a in-depth understanding of the conceptual meaning of combinations and the proof of the formula of it** .

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