What is the minimum number of friends Sally can have if she can invite a different group of friends to her house every night for a year?

Question

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?

Answer 1

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 $\binom{15}{3}=455$.

Answer 2

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.