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Mathematics 12 Online
OpenStudy (anonymous):

can u give a mathematical proof for C(n,r) = C(n,n-r)

OpenStudy (anonymous):

Here is the intuition behind it (though it is not a proof): A selection of 10 winners from a group of 17 is simultaneously a selection of 7 losers.

Parth (parthkohli):

\[C(n,r) = {n! \over (n - r)!r!} \]\[C(n,n - r) = {n! \over (n - (n - r))!(n-r)!}\]

Parth (parthkohli):

Simplify the second line.

OpenStudy (anonymous):

yeah i know that proof but not the mathematical one

Parth (parthkohli):

\[ C(n, n - r) = {n! \over (n - n + r)!(n - r)!} \]\[ \color{Black}{\Rightarrow {n! \over (r)!(n - r)!} }\]Which just gives the first line.

Parth (parthkohli):

It is mentioned to give a mathematical proof, isn't it?

OpenStudy (anonymous):

yeah

Parth (parthkohli):

So that's my response I gave you.

OpenStudy (anonymous):

number of ways of selecting r things from n things = number of ways of rejecting n-r things from n things....

OpenStudy (anonymous):

yeah thanks got it now

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