Can someone help me with the following:
see the attachment.
I am really stuck and do not know were to start
Thanks
Maths_Brain
Having been stupid enough to open said attachment what do I find? It's not a word document its a wrapper for picture!! Grrrr...
CB
(I'm not vulnerable to macro virus's but I don't like opening word and excel attachments because it takes a long time to boot up OOO on my slow computers (I won't open an office document on the my work computers as a matter of principle)
The third part is just algebra, write out the right hand side in terms of factorials and simplify.
$\displaystyle
{n-1 \choose r}+{n-1 \choose r-1}=\frac{(n-1)!}{r!(n-1-r)!} + \frac{(n-1)!}{(r-1)!(n-r)!}
$
Alternativly one may observe that it is the sum of the number of ways of choosing a subset of $\displaystyle r$ elements from the given set which does not include $\displaystyle 1$, and the number of ways that does include $\displaystyle 1$.
CB
When choosing a subset consisting of $\displaystyle r$ elements from a set of $\displaystyle n$ elements the first element may be choosen in $\displaystyle n$ ways the second in $\displaystyle (n-1)$ ways and so on to the $\displaystyle r$-th which may be choosen in $\displaystyle (n-r+1)$ ways. So the number of ways that the set may be choosen is $\displaystyle n(n-1)...(n-r+1)$, but in this number every permutation of each distinct choice appears, so to get the number of choices independent of order we divide by the number of permutations of $\displaystyle r$ distinct objects. Hence the number of ways of choosing the subset is:
$\displaystyle N(r,n)=\frac{n(n-1)...(n-r+1)}{r!}=\frac{n!}{r!(n-r)!}$
CB
When choosing a subset consisting of $\displaystyle r$ elements from a set of $\displaystyle n$ elements the first element may be choosen in $\displaystyle n$ ways the second in $\displaystyle (n-1)$ ways and so on to the $\displaystyle r$-th which may be choosen in $\displaystyle (n-r+1)$ ways. So the number of ways that the set may be choosen is $\displaystyle n(n-1)...(n-r+1)$, but in this number every permutation of each distinct choice appears, so to get the number of choices independent of order we divide by the number of permutations of $\displaystyle r$ distinct objects. Hence the number of ways of choosing the subset is:
$\displaystyle N(r,n)=\frac{n(n-1)...(n-r+1)}{r!}=\frac{n!}{r!(n-r)!}$
Which is what we expect.
Now the number of ways of choosing a subset that does not contain $\displaystyle 1$ is $\displaystyle N(r,n-1)$, and so the probability required is:
$\displaystyle p=\frac{N(r,n-1)}{N(r,n)}$
Similarly if $\displaystyle 1$ is in the set the number of ways of choosing it is $\displaystyle N((r-1),(n-1))$
CB