Henry Ernest Dudeney/Modern Puzzles/167 - A General Election/Solution

Modern Puzzles by Henry Ernest Dudeney: $167$

A General Election

In how many different ways may a Parliament of $615$ members be elected if there are only $4$ parties:

Conservatives, Liberals, Socialists, and Independents?

You see you might have $\text C. 310$, $\text L. 152$, $\text S. 150$, $\text I. 3$;

or $\text C. 0$, $\text L. 0$, $\text S. 0$, $\text I. 615$;

or $\text C. 205$, $\text L. 205$, $\text S. 205$, $\text I. 0$; and so on.

The candidates are indistinguishable, as we are only concerned with the party numbers.

Solution

$39 \, 147 \, 416$ different ways.

Proof

According to Dudeney:

The general solution is as follows.

Let $p =$ parties and $m =$ members.

Then $C^{p - 1}_{m + p - 1} =$ number of ways.

In the above, $C^{p - 1}_{m + p - 1}$ is the binomial coefficient $\dbinom {p - 1} {m + p - 1}$.

This theorem requires a proof. In particular: The above statement can be made abstract by invoking a theorem Number of Integer Partitions into Specific Number of Parts, or something like that. I think this, or at least something related, may be in Polya and Szego, which I promise one day I will embark upon. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $167$. -- A General Election
• 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $446$. A General Election