Question

At an election, a voter may vote for any number of candidates not greater than the number to be elected. There are 10 candidates and 4 are to be elected. The number of ways in which a voter may vote for at least one candidate is:
A.$385$
B.$1110$
C.$5040$
D.None of these

Answer 1

Hint: We will use the formula of combinations: ${}^n{C_r}$, where n is the total number of candidates and r is the number of candidates to be chosen, to find the number of ways in which a voter may vote for at least one candidate. Now, at least one candidate is to be voted, that means a voter can vote for more than 1 candidate. So, a voter can vote for only 1 candidate, 2, 3 or all 4 candidates. Hence, we will add all the ways in which a voter may vote for at least one candidate.

Complete step-by-step answer:
 We are given that a voter may vote for any number of candidates not greater than the number to be elected.
In total, there are 10 candidates and only 4 are to be elected.
Hence, by the given statement, a voter can vote for a maximum of 4 candidates.
We are required to calculate the number of ways in which a voter may vote for at least one candidate.
We know the number of ways can be calculated using the method of combinations given by: ${}^n{C_r}$ where C is the number of combinations, n is the total number of objects in the set and r is the number of objects to be chosen from the set.
Now, we are asked the number of ways in which a voter may vote for at least one candidate i.e, a voter may vote for more than 1 candidate. Also, a voter can vote a maximum of 4 candidates. So that means a voter can vote for either 1 candidate, 2 candidates, 3 candidates or all 4 candidates.
Therefore, the number of ways can be defined as: ${}^n{C_1} + {}^n{C_2} + {}^n{C_3} + {}^n{C_4}$
Here, n = 10 so we can write this equation as:
The total number of ways in which a voter may vote for at least one candidate is \[{}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + {}^{10}{C_4}\]
We know that ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ , so using this to solve the above expression, we get
$ \Rightarrow $Total number of ways: \[{}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + {}^{10}{C_4} = \]\[\dfrac{{10!}}{{\left( {10 - 1} \right)!1!}} + \dfrac{{10!}}{{\left( {10 - 2} \right)!2!}} + \dfrac{{10!}}{{\left( {10 - 3} \right)!3!}} + \dfrac{{10!}}{{\left( {10 - 4} \right)!4!}}\]
Using the expansion of$n! = \left( n \right)\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)..\left( 3 \right)\left( 2 \right)\left( 1 \right)$, we can write that
 $ \Rightarrow $Total number of ways = $\dfrac{{\left( {10} \right)9!}}{{9!1!}} + \dfrac{{\left( {10} \right)\left( 9 \right)8!}}{{8!2!}} + \dfrac{{\left( {10} \right)\left( 9 \right)\left( 8 \right)7!}}{{7!3!}} + \dfrac{{\left( {10} \right)\left( 9 \right)\left( 8 \right)\left( 7 \right)6!}}{{6!4!}}$
$ \Rightarrow $Total number of ways = $\dfrac{{10}}{1} + \dfrac{{90}}{2} + \dfrac{{720}}{6} + \dfrac{{5040}}{{24}}$
 = 10 + 45 + 120 + 210
 = 385
Hence, there are a total 385 ways in which a voter may vote for at least one candidate.
Therefore, option (A) is correct.

Note: In this question, you may get confused in the concept of voting “at least one” candidate because when we say at least one, it means we can have more than 1 but a minimum of 1 is necessary. We have used the formula of combinations and a combination is a method for determining the number of possible arrangements for a particular set of things or items (order is not mandatory).