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Proof

(related to Proposition: Simple Binomial Identities)

Ad \((1)\)

• From the closed formula for binomial coefficients and the definition of rational numbers it follows for \(k\ge 1\): \[\begin{array}{rcll} k\binom nk&=&\cancel {k}\cdot \frac{n(n-1)\cdot\ldots\cdot(n-k+1)}{\cancel {k}(k-1)\cdot\ldots \cdot 2\cdot 1}&\text{definition of binomial coefficients, cancellation of }k\\ &=&n\cdot \frac{(n-1)\cdot \ldots\cdot (n-k+1)}{(k-1)\cdot\ldots \cdot 2\cdot 1}&\text{extraction of a factor from the nominator}\\ &=&n\binom{n-1}{k-1}&\text{definition of binomial coefficients}. \end{array}\]

Ad \((2)\)

• From the closed formula for binomial coefficients, the definition of rational numbers, and the distributivity law for rational numbers, it follows for \(k\ge 2\):

\[\begin{array}{rcll} k^2\binom nk&=&(k+k(k-1))\binom nk&\text{rewriting }k^2\\ &=&k\binom nk+k(k-1)\binom nk&\text{"distributivity law for rational numbers}\\ &=&k\binom nk+\cancel {k(k-1)}\cdot \frac{n(n-1)(n-2)\cdot\ldots\cdot(n-k+1)}{\cancel {k(k-1)}(k-2)\cdot\ldots \cdot 2\cdot 1}&\text{definition of binomial coefficients, cancellation of }k(k-1)\\ &=&k\binom nk+n(n-1)\cdot \frac{(n-2)\cdot \ldots\cdot (n-k+1)}{(k-2)\cdot\ldots \cdot 2\cdot 1}&\text{extraction of a factor from the nominator}\\ &=&k\binom nk+n(n-1)\binom{n-2}{k-2}&\text{definition of binomial coefficients}. \end{array}\]

Ad $(3)$

• By the closed formula for binomial coefficients $$\binom nk=\frac{n !}{k !\cdot (n-k) !}.$$
• The formula is symmetric by replacing $k$ by $(n-k)$.
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References

Bibliography

1. Bosch, Karl: "Elementare Einführung in die Wahrscheinlichkeitsrechnung", vieweg Studium, 1995, 6th Edition