Question 1202888: In a town election, ten names of candidates were listed on a proportional representation ballot. The voters should place a 1 by the candidate of their first choice, 2 by their second choice, etc. until all ten candidates have numbers by their names. Voters must indicate their first choice, and others may be marked or not as they choose.
Determine how many different ways the voter may mark the ballot.
Not sure how to solve.
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13209)   (Show Source):
You can put this solution on YOUR website!
The voter has 10 choices for 1st, then 9 choices left for second, ..., and finally only 1 choice left for last.
Then the fundamental counting principle says that the total number of ways of marking the ballot is found by multiplying the numbers of choices for each one.
ANSWER: 10*9*8*...*2*1 = 10! = 3628800
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(added later, after tutor @ikleyn revised her response....)
Indeed the wording of the problem is terrible. Looking at it again, it is probable that her revised interpretation of the problem is correct -- that a first choice must be shown, and any further ranking is optional.
The statement of the problem could easily have been changed to make that clear.
Answer by ikleyn(52879)  (Show Source):
You can put this solution on YOUR website! .
In a town election, ten names of candidates were listed on a proportional representation ballot.
The voters should place a 1 by the candidate of their first choice, 2 by their second choice,
etc. until all ten candidates have numbers by their names.
Voters must indicate their first choice, and others may be marked or not as they choose.
Determine how many different ways the voter may mark the ballot.
Not sure how to solve.
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The wording of the problem is terrible.
The correct wording is this:
For any integer value of "k" between 1 and 10 inclusive,
the voter can mark "k" first positions,
leaving the rest of positions empty.
I changed and re-wrote my original solution.
Now you see here my changed final version.
10 possible options for one position filled.
10*9 = 90 possible options for first two positions filled.
10*9*8 = 720 possible options for first three positions filled.
10*9*8*7 = 5040 possible options for first four positions filled.
10*9*8*7*6 = 30240 possible options for first five positions filled.
10*9*8*7*6*5 = 151200 possible options for first six positions filled.
10*9*8*7*6*5*4 = 604800 possible options for first seven positions filled.
10*9*8*7*6*5*4*3 = 1814400 possible options for first eight positions filled.
10*9*8*7*6*5*4*3*2 = 3628800 possible options for first nine positions filled.
10*9*8*7*6*5*4*3*2*1 = 3628800 possible options for all ten positions filled.
Now the answer is the sum of all these partial products
10 + 90 + 720 + 5040 + 30240 + 151200 + 604800 + 1814400 + 3628800 + 3628800 = 9864100.
ANSWER. In 9864100 different ways, in all.
Solved.
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