Henry Ernest Dudeney/Puzzles and Curious Problems/120 - Proportional Representation/Solution

by : $120$

 * Proportional Representation
 * In a local election, there were ten names of candidates on a proportional representation ballot paper.
 * Voters should place No. $1$ against the candidate of their first choice.
 * They might also place No. $2$ against the candidate of their second choice,
 * and so on until all the ten candidates have numbers placed against their names.
 * The voters must mark their first choice, and any others may be marked or not as they wish.


 * How many different ways might the ballot paper be marked by the voter?

Solutions

 * $9 \, 684 \, 100$

Proof
Let $n$ denoted the number of different ways to vote.

Let the voter vote for $k$ candidates, where $1 \le k \le 10$.

Each way of making $k$ votes is an $n$-permutation on $10$ objects, denoted ${}^k P_{10}$.

Thus:
 * $n = \ds \sum_{k \mathop = 1}^{10} {}^k P_{10}$

where from Number of Permutations:


 * ${}^k P_{10} = \dfrac {10!} {\paren {10 - k}!}$

Hence we can calculate: