Club Election Voting Permutation Question

Question

In a club election the number of contestants is one more than the number of maximum candidates for which a voter can vote. If the total number of ways is which a voter can vote be 62, then the number of candidates is:

My Approach!
Let's suppose we have m voters and n be the maximum candidate so total number of contestants becomes n+1. 
Now each voter can vote in n+1 ways. So m voters can vote in m* (n+1) ways which is equal to 62.

That's it, I am not able to figure it out more.

Ramaz · Answer

Let there be $n$ candidates (contestants), and I assume that you have to vote for at least one of them, then 
the number of ways of casting votes is $binom{n}{1} + binom{n}{2} + dotsc + binom{n}{n-1}$.
[ Remember, you can only vote for a maximum of n-1 ]

Now it is known that the sum of the binomial coefficients, 
$$binom{n}{0} + binom{n}{1} + dotsc + binom{n}{n} = 2^n$$
subtracting $binom{n}{0} + binom{n}{n} = 1 + 1$, 
the number of ways to vote is $2^n - 2 = 62$
begin{gather}
2^n = 64 \
2^n = 2^6 \
n = 6
end{gather}

Especially Lime · Answer

It sounds like the only restrictions on who you can vote for are that you can't vote for everybody, and you can't vote for nobody. So you can vote for any subset of the $n$ candidates except those two. So you have $2^n-2$ different ways to vote, and so $2^n=64$.

Answered by Especially Lime on December 23, 2021

Club Election Voting Permutation Question

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