Prove the lecturer is a liar... I was given this puzzle:


  At the end of the seminar, the lecturer waited outside to greet the attendees. The first three seen leaving were all women. The lecturer noted " assuming the attendees are leaving in random order, the probability of that is precisely 1/3." Show the lecturer is lying (or badly mistaken).


I've puzzled it out to proving that there is no ratio of $\binom{a}{3}/\binom{a+b}{3}$ that is 1/3, where $ a,b
\in\mathbb{N}$ and $a\ge3$ and $b\ge0$, $a$ being the number of women and $b$ the number of men.

I'm stuck at this point (but empirically pretty convinced).

Any help/pointers appreciated.

Rasher

PS- as an amusing aside, the first 12 values in the sequence of values for $\binom{3+b}{3}$ are the total number of gifts received for each day of the "12 days of Christmas" song.

I've narrowed it down to proving that in the sequence generated by $n^3+3 n^2+2 n$ with $n
\in\mathbb{N}$ and $n\ge1$ it is impossible for $3(n^3+3 n^2+2 n)$ to exist in the form of $n^3+3 n^2+2 n$ . Still stymied at this point.

I found today a (somewhat) similar question at MathOverflow. Since my question seems to boil down to showing the Diophantine $6 a - 9 a^2 + 3 a^3 - 2 b + 3 b^2 - b^3=0$ has no solutions for $(a,b)
\in\mathbb{N}$ and $(a,b)&gt;= 3$ would it be appropriate to close this here and ask for help at MathOverflow to determine if this can be proved?

An update: I asked a post-doc here at Stanford if he'd have a look (he's done some heavy lifting in the area of bounds on ways $t$ can be represented as a binomial coefficient). To paraphrase his response "That's hard...probably beyond proof in the general case". Since I've tested for explicit solutions to beyond 100M, I'm settling with the lecturer is lying/mistaken at least in spirit unless one admits lecture halls the size of a state.
Let $a$ = the number of women, $b$ = the number of men, and $n = a + b$ be the total number of attendees.

The probability that the first 3 students to leave are all female is $\frac{a}{n} \cdot \frac{a-1}{n-1} \cdot \frac{a-2}{n-2}$.  Setting this expression equal to $\frac{1}{3}$ and cross-multiplying gives $3a(a-1)(a-2)=n(n-1)(n-2)$.

The product of any three consecutive integers is divisible by 6, so the left-hand side is divisible by 18.  For the equation to work out, we must have $n \in \{0, 1, 2\}$ modulo 9.

This doesn't solve your puzzle, but it does rule out (informally) 2/3 of the domain.
Your ratio is $\frac{a! (a+b-3)! 3!}{(a-3)!3!(a+b)!} = \frac{(a+b-3)!}{(a-3)!(a+b)!} = 1/\binom{a+b-3}{a+b},$ so you are trying to prove that $\binom{a+b}{a+b-3}$ is never equal to $3.$ If you look at Pascal's triangle, you will see that the third column is increasing, so you need only check it for a couple of initial values.