How many different groups of $4$ can be made from $142$? I performed this weekend in what we call extreme quarteting.  We had $142$ people and we wanted to know how many different groups of $4$ can $142$ people make?
 A: The answer is:
$$
{ C }_{ 142 }^{ 4 }\quad =\quad \frac { 142! }{ 4!(142-4)! } 
$$
Explanation:
When you choose the first eleent of the group you have 142 possibilities, for the second you'll have 141 because you have already chosen one, and for the third you'll have 140 following the same logic, and finally for the fourth i's 139 possibilities. Now using the principle of multiplication in counting you get:
$$
142\times 141\times 140\times 139\quad =\quad \frac { 142! }{ (142-4)! } 
$$
But this doesn't give us different groups, instead it gives all the possible arrangement. But we don't care about the ranking so we should decrease the number of possibilities.
Let's calculate it:
The number of possible ways of rearranging of a sequence made up of 4 different elements $ABCD$ is: $\frac{4!}{1!1!1!1!} = 4!$. Why?
Use the same logic we used for the previous answer except that here we care about arrangements so we can use it later to obtain our answer.
So our final answer would be :
$$
\frac { 142! }{ 4!(142-4)! } 
$$
A: In general, there are $\binom{n}{k}$ ways to form groups with $k$ people from $n$ persons, see here, so there are $\binom{142}{4}=16234505$ possibilities.
Useless remark: As abstract groups with $4$ elements there are only two different groups - the cyclic one and the Kleinian $4$-group.
