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The question: Sally has $N$ friends and likes to invite them over in small groups for dinner. She calculated that she can invite a different group of $3$ friends to dinner at her house every night for a year. What is the minimum number of friends Sally can have?

Use $365$ for days in a year; leap-year does not have to be considered.

We know $k=3$, so $\binom{n}{3}=\frac{n!}{3!(n-3)!}=\frac{n!}{6(n-3)!}=365$

$\frac{n!}{(n-3)!}=2190$

I get stuck at $n!=2190(n-3)!$

I assume that this is the next step in the process. What do I do from here?

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    Expanding on what hoyland said, $\frac{n!}{(n-3)!}=n(n-1)(n-2)$ giving you a cubic equation. You should also keep in mind that you are looking for the smallest $n$ such that $\binom{n}{3}\geq 365$, as it is unlikely that any integer will give you equality.2012-02-21

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I must admit I don't fully understand the conventions of the site for when we leave answers and when we give hints. So, this is a full solution.

So if Sally has $n$ friends, there are $\binom{n}{3}$ ways to invite them over. So, as you said, set this equal to 365. $\binom{n}{3}=365$ $\dfrac{n!}{(n-3)!6}=365$ $\dfrac{n!}{(n-3)!}=2190$ But recall what $n!$ means:

$\dfrac{n(n-1)\cdots(2)(1)}{(n-3)(n-4)\cdots(2)(1)}=2190$ So we can cancel:

$(n)(n-1)(n-2)=2190$

At this point, we have a cubic: $n^3-3n^2+2n-2190$. It has one real root, roughly $14.01$. Since we need $n$ to be an integer, Sally needs 15 friends.

We can check: $\binom{14}{3}=364$ (so close!) and $\binom{15}{3}=455$.

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    Thanks for explaining the step between n!/(n-3)!=2190 and (n)(n-1)(n-2)=21902012-02-21
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To start with, you should be using an inequality. What you actually have is $\frac{n!}{(n-3)!} \ge 2190$

You can simplify the LHS to get $n(n-1)(n-2) \ge 2190$

Now, since $n$ is positive, $(n-2)^3 \le n(n-1)(n-2) \le n^3$, so observing that $\sqrt[3]{2190} \approx 12.9861786$ we can do trial and error starting at $n=13$. It turns out that $\binom{14}{3} = 364$, so the answer is that she has at least 15 friends.