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Pascal’s Triangle and Binomial Coefficients.md

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