At an election, a voter may vote for any number of candidates, not greater than the number to be ele

Q: Wait, why can't I just use $2^{10} - 1$? Doesn't that count all possible non-empty subsets?

That is a brilliant observation, but it is a trap! The formula $2^{n} - 1$ counts *all* possible non-empty subsets of candidates, which includes voting for 5, 6, 7, 8, 9, or even all 10 candidates. The problem explicitly restricts you to a maximum of 4. We must only sum the cases where $r \le 4$.

Q: How do I know when to use Permutations versus Combinations?

Ask yourself: 'If I swap the order of my choices, does the outcome change?' In a ballot, voting for Candidate A then Candidate B is exactly the same as voting for Candidate B then Candidate A. Since the order of selection does not change the outcome, we use Combinations.

Q: Why do we add the cases instead of multiplying them?

We add because these are mutually exclusive scenarios. You cannot vote for exactly 1 candidate AND exactly 2 candidates at the same time on a single ballot. In combinatorics, when you have distinct, non-overlapping scenarios (the 'OR' condition), you add the possibilities.

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 of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is

Options:

(A) 5040

(B) 6210

(C) 385

(D) 1110

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are walking into a polling booth. You look at the ballot paper, and right in front of you, there are ten different candidates standing for the election.

In any election problem, the first step is to identify our total pool of options. Here, our total number of candidates, which we will call $n$, is exactly $10$.

The election is only to select four people. That means, out of these ten candidates, a maximum of four can actually win and be elected. We can visualize this as four empty slots waiting to be filled on your ballot paper.

### Defining the Constraints

The problem states two very important conditions that we must satisfy simultaneously. First, you must vote for at least one candidate; you cannot just drop a blank ballot into the box.

Second, you cannot vote for more candidates than the number to be elected. So, the number of votes you cast must be strictly between one and four, inclusive.

Because of these strict constraints, we can break our complex problem down into four distinct, mutually exclusive cases: choosing exactly one, two, three, or four candidates. Breaking a problem into cases is a very powerful technique in combinatorics.

### The Combinatorial Tool

Before we start calculating, ask yourself a fundamental question: does the order in which you tick the names on the ballot matter? No, it absolutely does not.

Since the order of selection does not matter, this is a classic combinations problem. We will use our trusty formula:

$$^{n}C_{r} = \frac{n!}{r!(n-r)!}$$

### The Calculation Phase

Let's tackle Case 1: voting for just one single candidate. Out of the ten available candidates, you need to choose exactly one.

Mathematically, we write this as $^{10}C_{1}$. Since any number $^{n}C_{1}$ is simply $n$, there are exactly $10$ ways to cast a vote for one candidate.

Moving on to Case 2: supporting two candidates. You are selecting two individuals out of the pool of ten, expressed as $^{10}C_{2}$.

$$^{10}C_{2} = \frac{10 	imes 9}{2 	imes 1} = 45 	ext{ ways}$$

Now let's look at Case 3: voting for three candidates. Choosing three from ten is written as $^{10}C_{3}$.

$$^{10}C_{3} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120 	ext{ ways}$$

Finally, we reach Case 4: voting for four candidates, the maximum allowed. This is $^{10}C_{4}$.

$$^{10}C_{4} = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210 	ext{ ways}$$

### The Grand Summation

We have successfully found the number of ways for each individual case. Because these are mutually exclusive events (you choose one OR two OR three OR four), we must add these possibilities together.

The final addition equation is:

$$	ext{Total Ways} = ^{10}C_{1} + ^{10}C_{2} + ^{10}C_{3} + ^{10}C_{4}$$

Let's bring it all together for the grand finale:

$$	ext{Total Ways} = 10 + 45 + 120 + 210 = 385$$

You have successfully navigated the constraints, identified the correct combinatorial tool, and summed the possibilities. The final answer is **385 ways**.

Select Answer:

Visualized Solution

Identifying the Total Options

Understanding the Maximum Limit

Analyzing the Constraints

Breaking into Mutually Exclusive Cases

The Mathematical Tool: Combinations

Calculating Case 1

Calculating Case 2

Calculating Case 3

Calculating Case 4

Setting Up the Final Summation

Final Calculation

The Sigma Insight: Combinations and Selection

Analyzing the Setup

Defining the Constraints

The Combinatorial Tool

The Calculation Phase

The Grand Summation

Similar Questions

A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is

A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is

Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen?

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is _____

The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is :

The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is :