Question

(Probability) Combinations calculations (nCr) help/tips?

Is there a fast way to compute the following equalities? The first one isn't too bad, but I don't know how to do the 2nd one.. or why it works? (It seems you're just adding the terms but can you explain why this is true?)
\[x*\left(\begin{matrix}20 \\ x\end{matrix}\right)=20*\left(\begin{matrix}19 \\ x-1\end{matrix}\right)\] (Probability) Combinations calculations (nCr) help/tips?

Is there a fast way to compute the following equalities? The first one isn't too bad, but I don't know how to do the 2nd one.. or why it works? (It seems you're just adding the terms but can you explain why this is true?)
\[x*\left(\begin{matrix}20 \\ x\end{matrix}\right)=20*\left(\begin{matrix}19 \\ x-1\end{matrix}\right)\] @Mathematics

Answer 1

Is there a fast way to compute the following equalities? The first one isn't too bad, but I don't know how to do the 2nd one.. or why it works? (It seems you're just adding the terms but can you explain why this is true?)
\[x*\left(\begin{matrix}20 \\ x\end{matrix}\right)=20*\left(\begin{matrix}19 \\ x-1\end{matrix}\right)\]
\[\left(\begin{matrix}19 \\ x-1\end{matrix}\right)\left(\begin{matrix}60 \\ 10-x\end{matrix}\right)=\left(\begin{matrix}19 \\ x-1\end{matrix}\right)\left(\begin{matrix}60 \\ 9-(x-1)\end{matrix}\right)=\left(\begin{matrix}79 \\ 9\end{matrix}\right)\]

Answer 2

Oops i forgot to write there's the sigma sign from 0 to 10 for the 2nd one

Answer 3

\[\sum_{x=0}^{10}\left(\begin{matrix}19 \\ x-1\end{matrix}\right)\left(\begin{matrix}60 \\ 9-(x-1)\end{matrix}\right)\]

Answer 4

\[x*\frac{20(20-1)(20-2)...(20-x)}{x(x-1)(x-2)...(2)(1)}\]

\[=\frac{20(20-1)(20-2)...(20-x)}{(x-1)(x-2)...(2)(1)}\]

\[=20*\frac{(20-1)(20-2)...(20-x)}{(x-1)(x-2)...(2)(1)}\]

Answer 5

I'd just keep it with the factorials

\[x{20\choose x}=x\frac{20!}{x!(20-x)!}=\frac{20!}{(x-1)!(20-x)!}\]

\[20\frac{19!}{(x-1)!(20-x)!}=20\frac{19!}{(x-1)!(19-(x-1))!}=20{19\choose x-1}\]

Answer 6

also as a \[LaTeX\] tip use { n\choose r}

for \[{ n\choose r}\]

Answer 7

\[\sum_{x=0}^{10}\left(\begin{matrix}19 \\ x-1\end{matrix}\right)\left(\begin{matrix}60 \\ 9-(x-1)\end{matrix}\right)=205811513765\]

Answer 8

0-1 = -1; i havent come across factorials of negatives yet ...

Answer 9

it gives zero \[{19\choose -1}=0\]

Answer 10

how do we find that sum, without entering the sum for each term one by one from 0 to 10? since they have 79 Choose 9 i figured they used some formula?

Answer 11

this is the numerator of the hypergeometric distribution

Answer 12

\[\sum_{x=0}^m\frac{{n\choose x}{N-n\choose m-x}}{{N\choose m}}=1\]

thus

\[\sum_{x=0}^m{n\choose x}{N-n\choose m-x}={N\choose m}\]

Answer 13

ohh i see now! ah thank you so much!

Answer 14

no problem