Summation from k to m of r Choose k by s Choose n-k by nr-(r+s)k
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Theorem
- $\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$
Proof
The proof proceeds by induction over $m$.
For all $m \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
- $\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$
Basis for the Induction
$\map P 0$ is the case:
\(\ds \sum_{k \mathop = 0}^0 \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k}\) | \(=\) | \(\ds \dbinom r 0 \dbinom s {n - 0} \paren {n r - \paren {r + s} 0}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom s n \paren {n r}\) | Binomial Coefficient with Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {0 + 1} \paren {n - 0} \dbinom r {0 + 1} \dbinom s {n - 0}\) | Binomial Coefficient with One |
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P j$ is true, where $j \ge 1$, then it logically follows that $\map P {j + 1}$ is true.
So this is the induction hypothesis:
- $\ds \sum_{k \mathop = 0}^j \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {j + 1} \paren {n - j} \dbinom r {j + 1} \dbinom s {n - j}$
from which it is to be shown that:
- $\ds \sum_{k \mathop = 0}^{j + 1} \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {j + 2} \paren {n - j - 1} \dbinom r {j + 2} \dbinom s {n - j - 1}$
Induction Step
This is the induction step:
\(\ds \) | \(\) | \(\ds \sum_{k \mathop = 0}^{j + 1} \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^j \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} + \dbinom r {j + 1} \dbinom s {n - j - 1} \paren {n r - \paren {r + s} \paren {j + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {j + 1} \paren {n - j} \dbinom r {j + 1} \dbinom s {n - j} + \dbinom r {j + 1} \dbinom s {n - j - 1} \paren {n r - \paren {r + s} \paren {j + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom r {j + 1} \paren {\paren {j + 1} \paren {n - j} \dbinom s {n - j} + \dbinom s {n - j - 1} \paren {n r - \paren {r + s} \paren {j + 1} } }\) | factorising | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {j + 2} \paren {r - j + 2} \dbinom r {j + 2} \paren {\paren {j + 1} \paren {n - j} \dbinom s {n - j} + \dbinom s {n - j - 1} \paren {n r - \paren {r + s} \paren {j + 1} } }\) | Factors of Binomial Coefficient: Corollary $2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {j + 2} \paren {r - j + 2} \dbinom r {j + 2} \paren {\dfrac {\paren {j + 1} \paren {n - j} } {\paren {n - j} \paren {s - \paren {n - j} } } \dbinom s {n - j - 1} + \dbinom s {n - j - 1} \paren {n r - \paren {r + s} \paren {j + 1} } }\) | Factors of Binomial Coefficient: Corollary $2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {j + 2} \paren {r - j + 2} \dbinom r {j + 2} \dbinom s {n - j - 1} \paren {\dfrac {j + 1} {s - n + j} + \paren {n r - \paren {r + s} \paren {j + 1} } }\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {j + 2} \paren {r - j + 2} \dbinom r {j + 2} \dbinom s {n - j - 1} \paren {\dfrac {j + 1} {s - n + j} + \paren {n r - r j - s j - r - s} }\) | simplifying |
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So $\map P j \implies \map P {j + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $53 \ \text{(a)}$