There's a party and there are people clinking glasses. We hear 28 clinks (one person clinks with exactly one person). I have to find out how many people there are at the party.
So I put in this equation:
$\binom{x}{2} = 28$, but I don't know how to find $x$.
On
You can solve for $x$ in the polynomial $$28=\binom{x}{2}=\frac{1}{2}x(x-1)$$ as others mention.
However, 28 is small, and $x$ is limited to a positive integer. I'd use trial and error.
When $x=4$ we have $\binom{x}{2}=6$, which is too small.
When $x=5$ we have $\binom{x}{2}=10$, which is too small.
When $x=6$ we have $\binom{x}{2}=15$, which is too small.
...and keep going until you find the $x$-value which satisfies $\binom{x}{2}=28$.
On
The useful form of the binomial coefficient here $\binom x2 = \frac{(x)_2}{2!}$ uses the falling factorial, $(x)_2=x(x-1)$ (note the two terms; also be aware alternative notations exist).
This gives us $\binom x2 = \frac{x(x-1)}{2}$. Observing that $28$ is divisible by $7$ we can trial both $x=7$ and $x-1=7$ as our first-try likely solutions; alternatively we can solve the resulting quadratic equation $x^2-x-56=0$.
Ethan Bolker proposes a variation in comments:
When I clink glasses in a group I suggest that beyond four people the graph of who-clinks-with-who just needs to be connected, not complete. How many people might be there if you heard 28 clinks?
Here the all-clink-all solution gives the minimum value for this, and as posed the maximum value is given by the minimally connected graph, a tree, which fortunately has exactly one less edge (clink) than node (person), for a maximum number of people of $29$ (example: clinking that is "transmitted" down the line).
However this doesn't quite mesh with the "beyond four people" concept which suggests that any one person would want to clink with at least three other people (a graph of minimum degree $3$). Then $28$ clinks would imply up to $\left\lfloor \frac {2\cdot 28}{3} \right\rfloor = 18$ people.
On
Shouldn't the answer be 56?
The question says that "one person clinks with exactly one person". Which I would assume means that a every person only clinks once. Using binomial theorem, we will be counting cases in which a person clinks multiple times. I feel that the answer sould be 56, as 56 people can be divided into 2 groups of 28 giving 28 clinks.
You are correct that the equation to look at is $\binom{x}{2}=28$ by the handshaking lemma.
Next, remember that $\binom{x}{2}=\frac{x(x-1)}{2}$ so the above equation can be rewritten as
$$x^2-x=56$$
From here you may rearrange a bit and use the quadratic formula to solve for $x$ or use other introductory methods to find $x$: