Problem 158 - A Complete Resource Book in Mathematics for JEE Main

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 elected. If a voter votes for at least one candidate, then the number of ways in which he can vote is (A) 5040 (B) 6210 (C) 385 (D) 1110

Step-by-Step Solution

Answer

(C) 385

Step 1: Determine the Total Ways to Vote

The voter can choose 1 to 4 candidates to vote for out of 10 candidates. This is a combination problem.The number of ways to choose 1 candidate from 10 is \( \binom{10}{1} = 10 \).The number of ways to choose 2 candidates from 10 is \( \binom{10}{2} = 45 \).The number of ways to choose 3 candidates from 10 is \( \binom{10}{3} = 120 \).The number of ways to choose 4 candidates from 10 is \( \binom{10}{4} = 210 \).

Step 2: Sum All Possible Combinations

Add up all the different ways a voter can vote for 1, 2, 3, or 4 candidates:\[\binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4} = 10 + 45 + 120 + 210 = 385\]

Step 3: Verify the Conditions

The problem specifies that the voter must vote for at least one candidate. Our calculation considers all possible combinations where the voter picks at least 1 candidate, fitting these conditions.

Key Concepts

Election Voting Problem

In an election voting problem, we are often faced with scenarios where voters choose candidates from a list of nominees. This specific concept explores the decision-making process of voters, especially when they are allowed to select more than one candidate. This method, known as approval voting, offers voters the ability to support more than one nominee, up to a specified limit.
In our case, there are 10 candidates and 4 positions available. The nature of this scenario requires an understanding of how voters might decide whom to vote for, considering they can vote for between 1 and 4 candidates. This flexibility in choosing makes it crucial to count all possible combinations of choices. Thus, understanding this election type is important in contexts like committees and panels, where more than one candidate can be elected. In such problems, combinatorics comes into play to count the possible outcomes of voter choices.

Combinatorics

Combinatorics is a branch of mathematics dealing with combinations and permutations. It is fundamental in situations where we need to count or list the possible arrangements of objects based on certain rules.
In our voting problem, combinatorics helps us determine how voters can choose candidates in various ways. The essence of the task is to calculate how many different sets of candidates a voter might select given the constraints. Such calculations are done using combinations, which consider the selection of items without regard to order. Here, we break it down:

• Choosing 1 candidate from 10: calculated using a combination formula.
• Choosing 2 candidates: involves another combination calculation.
• And so on for 3 and 4 candidates.

The ability to sum up these different choices gives us the total number of ways a voter can exercise their voting preference. This approach highlights the power of combinatorics in simplifying complex decision-making processes.

Binomial Coefficient

The binomial coefficient is a key tool in combinatorics that assists in calculating combinations. It is represented as \( \binom{n}{k} \), where \( n \) is the total number of items, and \( k \) is the number of items to be chosen. This formula is used extensively when solving problems like the election voting problem.
To compute \( \binom{10}{k} \) (for \( k = 1, 2, 3, 4 \)), we use the formula:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]This formula calculates the number of ways to choose \( k \) candidates out of 10. "!" denotes factorial, which is the product of all positive integers up to that number. So, for our problem, each combination calculation tells us how many ways a set number of candidates can be chosen from the entire pool of 10 candidates.
By summing these calculated coefficients for \( k = 1 \) to \( 4 \), we fulfill the condition of choosing at least one candidate and find out the total possible voting combinations. The sum results in 385 distinct ways a voter can make their selections, emphasizing the elegance and practical application of the binomial coefficient in solving real-world problems.