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Question: In an election, the number of candidates is 1 greater than the person to be elected. If a voter can ...

In an election, the number of candidates is 1 greater than the person to be elected. If a voter can vote 254 ways, the number of candidates is
(a) 7
(b) 10
(c) 8
(d) 6

Explanation

Solution

Hint: Here, we will assume the number of candidates as n and person to be elected as n−1n-1 . Then we can say that he/she can vote for any 1 person, or any 2 person, or any 3 person up-to n−1n-1 . He/she cannot vote to n candidates. So, from this series of combinations will be formed which will be equal to value 2n{{2}^{n}} and then this equation we have to equate it with 254. Thus, we will get the value of n by solving.

Complete step-by-step answer:
In the question, we are given that the number of candidates is 1 greater than the person to be elected. So, we will assume that the number of candidates to be n and the person to be elected as n−1n-1 .
Now, we can say that a voter can vote in 254 ways i.e. he/she can vote any 1 person, or any 2 person, or any 3 person up-to n−1n-1 . He/she cannot vote to n candidates.
So, this in mathematical form can be written as
nC1+nC2+nC3+.....nCn−1{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{3}}+.....{}^{n}{{C}_{n-1}}
Now, we will add two terms here i.e. nC0{}^{n}{{C}_{0}} and nCn{}^{n}{{C}_{n}} . Also, we will subtract this in series. So, we get
nC0+nC1+nC2+nC3+.....nCn−1+nCn−nC0−nCn{}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{3}}+.....{}^{n}{{C}_{n-1}}+{}^{n}{{C}_{n}}-{}^{n}{{C}_{0}}-{}^{n}{{C}_{n}}
Now, we know that nC0+nC1+nC2+nC3+.....nCn−1+nCn=2n{}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{3}}+.....{}^{n}{{C}_{n-1}}+{}^{n}{{C}_{n}}={{2}^{n}} and nC0,nCn=1{}^{n}{{C}_{0}},{}^{n}{{C}_{n}}=1 So, putting this value in the expression, we get
2n−1−1=2n−2{{2}^{n}}-1-1={{2}^{n}}-2
Now, it is given that there are a total 254 ways to elect the candidate. So, we can write it as
2n−2=254⇒2n=256{{2}^{n}}-2=254\Rightarrow {{2}^{n}}=256
We know that multiplying 2 eight times we get value 256 so, we get value of n as
28=256⇒n=8{{2}^{8}}=256\Rightarrow n=8
Thus, we have assumed the number of candidates as n which is 8.
Hence, option (c) is correct.

Note: Another approach is by assuming an elected person to be n so, the number of candidates will be 1 greater than n i.e. n+1n+1 . So, voters have 2 choices to elect the person. Thus, there are 2n+1{{2}^{n+1}} choices. Now, at least one candidate is to be selected and all candidates cannot be voted. So, we will subtract 2 from the choices, we will get as
2n+1−2=254{{2}^{n+1}}-2=254
On solving, we get
2n+1=256=28{{2}^{n+1}}=256={{2}^{8}}
n+1=8⇒n=7n+1=8\Rightarrow n=7
Thus, we get the number of elected people as 7 but we have assumed the number of candidates as n+1n+1 so, we get the answer as n=7+1=8n=7+1=8 . So, the answer will be the same by this method.