Henry Ernest Dudeney/Puzzles and Curious Problems/273 - City Luncheons/Solution

by : $273$

 * City Luncheons

Solution
There are $15$ members of staff at each firm.

Proof
Let $n$ be the number of staff at both Pilkins and Popinjay and Radson, Robson, and Ross.

The number of ways to seat $3$ people together out of $n$ is the binomial coefficient $\dbinom n 3 = \dfrac {n \paren {n - 1} \paren {n - 2} } {3 \times 2 \times 1}$.

The number of ways to seat $4$ people together out of $r$ is the binomial coefficient $\dbinom n 4 = \dfrac {n \paren {n - 1} \paren {n - 2} \paren {n - 3} } {4 \times 3 \times 2 \times 1}$.

Hence we want to find $n$ such that:
 * $3 \dfrac {n \paren {n - 1} \paren {n - 2} } {3 \times 2 \times 1} = \dfrac {n \paren {n - 1} \paren {n - 2} \paren {n - 3} } {4 \times 3 \times 2 \times 1}$

This simplifies down to:
 * $12 = n - 3$

from which the answer is that $n = 15$.

We see that:
 * $\dbinom {15} 3 = \dfrac {15 \times 14 \times 13} {3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$

and:
 * $\dbinom {15} 4 = \dfrac {15 \times 14 \times 13 \times 12} {4 \times 3 \times 2 \times 1} = 5 \times 7 \times 13 \times 3 = 1365 = 455 \times 3$