Ways to select 3 members from 5 candidates
At an election there are 5 candidates and 3 members are to be selected. In how many ways a voter can vote?
My attempt: 1st member can be chosen in 5 ways, 2nd in 4 and 3rd in 3. So, $5*4*3=60$.
But the answer is given as 25. And the hint says $C(5,1)[C(3,1)+C(2,1)].$ I am not getting it.
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$\begingroup$Your answer double counts possible permutations of the three selected candidates. For example, you are counting $$\quad A, B, C,\quad A, C, B,\quad B, A, C,\quad B, C, A,\quad C, A, B,\quad \text{and}\; C, B, A$$ as six separate possible choices, and yet they all represent precisely the same three candidates.
We could use your calculation, but then divide by $3! = 6$ to account for these permutations:
$$\dfrac{5\cdot 4 \cdot 3}{3!} = \dfrac{60}{6} = 10$$
Shorter yet, we have $5$-choose-$3$ ways a voter can select candidates: $$\binom{5}{3} = \frac{5!}{3!2!} = 10$$
In order to obtain an answer of $25$, you would need to assume that a voter has the option to vote for at least one candidate and at most three candidates.
$\endgroup$ 8 $\begingroup$Here is how to get 25 as result:
Assume that you have the option to vote for a maximum of 3 candidates. Then you have 10 = 5-choose-3 = 5*4*3 / 3! options for voting for exactly three candidates, 10 = 5-choose-2 = 5*4 / 2! for voting for two candidates, and 5 = 5-choose-1 options for voting for only one candidate. The grand total is 25 options for voting for at least one and at most 3 out of 5 candidates.
Unfortunately, I cannot make sense of the hint right away, so I can only offer the guess that there is another way to think of this problem. If I had to give a hint in this form, using the notation C(n,k) for n-choose-k, I'd have hinted C(5,3) + C(5,2) + C(5,1)
EDIT: Added 5-choose-k equalities as explanation and a paragraph about (how little my answer related to) the hint.
$\endgroup$ $\begingroup$The condition is that a voter can't vote for more than $3$ candidates. But, he is entitled to vote for even $1$ candidate. Thus, $5C1 + 5C2 + 5C3 = 25$. Why we add them is for the reason that, a voter can vote for one candidate or two candidates or three candidates. So we get $25$.
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