Question

At an election there are five candidates and three members to be elected, and an elector may vote for any number of candidates not greater than the number to be elected. Then the number of ways in which can elector may vote is:
(a) 25
(b) 30
(c) 32
(d) None of these

Answer 1

Hint: For solving this question you should know about the concept of permutation and combination. Combination is a way of selecting items from a collection where the order of selection does not matters. Suppose we have a set of three numbers A, B and C. Then in how many ways we can select two numbers from each set, is defined by combination.

Complete step-by-step solution:
According to the question at an election there are five candidates and three numbers to be elected and an elector may vote for any number of candidates not greater than the number of candidates not greater than the number to be elected. Then how many ways in which an elector may vote is.
So, we will solve this equation by combination method. If we understand the combination by an example then it will be clear.
For example, If we want to buy a milkshake and we are allowed to combine three flavours from Apple, Banana, cherry and papaya. Then the combination of Apple, Banana and cherry is the same as the combination of Banana, Apple and cherry. So, if we are supposed to make a combination out of these possible flavours, then firstly, we will shorten the name of the fruits by selecting the first letter of their name, And then the combination of four is ABC, ABP, ACP, BCP.
So, the total number of ways of 4 here.
And we can also calculate this by formula
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
So, according to our question: A voter can give either 1 vote, 2 vote or 3 vote.
So, Number of ways to give only 1 vote \[={}^{5}{{C}_{1}}=5\]
Number of ways to give only 2 votes \[={}^{5}{{C}_{2}}=10\]
Number of ways to give all three votes \[={}^{5}{{C}_{3}}=10\]
So, the voter can give vote as total \[=10+10+5=25\]

Note: If we want to solve these type questions then always use combination and this combination is used for selecting some events from complete events. When we can’t write all events there. So, we use it. And at some places we use permutation also.