+23246 nCr calculates the number of combinations of sub-groups of size 'r' taken from an initial group of size 'n'.
For instance, if there is a group of 20 persons and 4 are chosen to represent that group, there will be 20C4 possible groups.
When using combinations, order is not important -- there will be a group of 4 persons but the group of {Tom, Betty, Mary, Carl} is the same group as {Betty, Carl, Mary, Tom}.
It is calculated with this formula: n! / ( r! · (n - r)! ).
+23246 nCr calculates the number of combinations of sub-groups of size 'r' taken from an initial group of size 'n'.
For instance, if there is a group of 20 persons and 4 are chosen to represent that group, there will be 20C4 possible groups.
When using combinations, order is not important -- there will be a group of 4 persons but the group of {Tom, Betty, Mary, Carl} is the same group as {Betty, Carl, Mary, Tom}.
It is calculated with this formula: n! / ( r! · (n - r)! ).
+118609 Thanks Geno,
I'm just giving another example.
Say you have 5 b***s that are lettered as A,B,C, D and E
how many ways can you choose 3 or them
Let me see
ABC ABD ABE ACD ACE ADE
BCD BCE BDE
CDE
that is 10 ways. (assuming that I didn't double up or miss any)
5C3 is just this. It is how many ways you can select 3 things out of 5 things so it must equal 10
check
I'll enter nCr(5,2) into the web 2 calc. :)
$${\left({\frac{{\mathtt{5}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{10}}$$
Oh and if you are doing it ny hand the calc just showed you the formula :)
