44 lines
1.2 KiB
Markdown
44 lines
1.2 KiB
Markdown
#Math #Probability
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# Observing Pascal’s Triangle
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| n/k | 0 | 1 | 2 | 3 | 4 | 5 |
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| --- | --- | --- | --- | --- | --- | --- |
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| 0 | 1 | | | | | |
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| 1 | 1 | 1 | | | | |
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| 2 | 1 | 2 | 1 | | | |
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| 3 | 1 | 3 | 3 | 1 | | |
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| 4 | 1 | 4 | 6 | 4 | 1 | |
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| 5 | 1 | 5 | 10 | 10 | 5 | 1 |
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As you can see, Pascal’s Triangle generates:
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$$
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{n \choose k}
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$$
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or
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$$
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\frac{n!}{k!(n-k)!}
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$$
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But how does this work?
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First, we can manually prove the top two rows of Pascal’s Triangle by plugging the values into the binomial coefficient formula.
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Afterward, we can use the property of Pascal’s Triangle, taking Pascal’s Triangle as a function P:
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$$
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P(n + 1, k) = P(n, k) + P(n, k-1)
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$$
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By proving this property in the binomial coefficient formula, we can deduce that Pascal’s Triangle generates binomial coefficients
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# The Proof
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$$
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\frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-k+1)!}\\=\frac{n!(n-k+1)}{k!(n-k)!(n-k+1)}+\frac{n!k}{(k-1)!(n-k+1)!k}\\=\frac{n!(n-k+1)}{k!(n-k+1)!}+\frac{n!k}{k!(n-k+1)!}\\=\frac{n!(n+k+1-k)}{k!(n-k+1)!}\\=\frac{n!(n+1)}{k!(n-k+1)!}\\=\frac{(n+1)!}{k!(n-k+1)!}
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$$
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From this, we have proven that we can generate binomial coefficients using Pascal’s Triangle |