Question

In an election, the number of candidates excesses the number to be elected by 2. A man can vote 56 ways. Then, the number of candidates is
a. 5
b. 6
c. 7
d. 8

Answer 1

Hint: Suppose the number of total candidates as a variable. Now, use the relation$C_{r}^{n}$for selecting r things from n things. One man can elect one candidate or two or three or the maximum possible number of total elected people from the total candidates available. Now calculate the number of ways by a man for visiting and use given conditions.

Complete step-by-step answer:

Let us suppose the number of candidates in the election be ‘x’ and hence, the number of candidates to be elected would be ‘x-z’ as it is given in the problem that the number of candidates exceeds the number to be elected by 2, in the election.

Hence, a man can select any number of candidates from ‘x-2’ persons as ‘x-2’ numbers are the candidates who can be elected by ‘x’ candidates.

Now, we can think of the above statement by the concept of permutations and combinations. So, in other words, we can select 1 candidates out of ‘x-2’ candidates from x persons, 2 candidates out of (x – 2) candidates from x persons, 3 candidates from x candidates and so on ; hence, we can select maximum (x – 2) candidates from ‘x’ candidates as it is given that total candidates to be elected can be (x – 2) (maximum).

Now, we know that selection of ‘r’ things out of ‘n’ things can be given by relation$C_{r}^{n}$which identity can be given as
$C_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!}$
Where n! = 1.2.3……….n
And n > 0 and $r\ge 0$and $r\ge r$.

So, he can elect 1 candidate from ‘x’ candidates by $C_{1}^{x}$ , elect 2 candidates from ‘x’ candidates by $C_{2}^{x}$, elect ‘3’ by $C_{3}^{x}$ and so on, similarly maximum elected candidates i.e. x – 2 candidates, can be elected from x candidates by $C_{x-2}^{x}$ ways.

Now, a man can vote by adding all different ways of selection of elected candidates from total ‘x’ candidates. Hence, it can be given by relation:
= number of ways of selecting (1 candidate + 2 candidates + 3 candidates + ……………….. (x -2) candidates) from x candidates.
$=C_{1}^{x}+C_{2}^{x}+C_{3}^{x}+.............C_{x-2}^{x}$ ………………….. (i)

As we already know the total number of ways of voting by a man is 56 as given in the question. Hence, we can equate 56 and equation (i) as both are representing the same values.
Hence, we get
$=C_{1}^{x}+C_{2}^{x}+C_{3}^{x}+.............C_{x-2}^{x}=56$ ……………. (ii)

Now, we know the following relation from the binomial chapter as
$=C_{0}^{n}+C_{1}^{n}+C_{2}^{n}+.............C_{n}^{n}={{2}^{n}}$
Replace ‘n’ by ‘x’ in the above, relation; hence, we get
$=C_{0}^{x}+C_{1}^{x}+C_{2}^{x}+.............C_{x}^{x}={{2}^{x}}$……….. (iii)

We can write the equation (iii) as
$C_{0}^{x}+\left( C_{1}^{x}+C_{2}^{x}+.............C_{x-2}^{x} \right)+C_{x-1}^{x}+C_{x}^{x}={{2}^{x}}$
Now, put value of $C_{1}^{x}+C_{2}^{x}+.........C_{x-2}^{x}$ from the equation (ii); hence, we get
$C_{0}^{x}+56+C_{x-1}^{x}+C_{x}^{x}={{2}^{x}}$

Now, use the relation $C_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!}$in the above relation and 0! = 1 to simplify the above equation further. Hence, we get
$\dfrac{x!}{x!0!}+56+\dfrac{x!}{\left( x-1 \right)!1!}+\dfrac{x!}{x!0!}={{2}^{x}}$
$\Rightarrow 1+56+\dfrac{x\left( x-1 \right)!}{\left( x-1 \right)!}+1={{2}^{x}}$
$58+x={{2}^{x}}$
$\Rightarrow {{2}^{x}}-x=58$

As the above equation is not a polynomial, it’s a combination of algebraic and exponential functions. So, we can solve this equation by hit and trial method. So, put values of x from 1 and observe which value will satisfy and provide LHS = RHS.

Hence, put x = 1,
$LHS={{2}^{1}}-1=1\ne RHS$
Hence, it’s not a solution.

Put x = 2,
$LHS={{2}^{2}}-2=4-2=2\ne RHS$
So, 2 is also not a solution.

Put x = 3,
$LHS={{2}^{3}}-3=8-3=5\ne RHS$
So, 3 is also not a solution.

Put x = 4,
$LHS={{2}^{4}}-4=16-4=12\ne RHS$
So, 4 is also not a solution.

Put x = 5,
$LHS={{2}^{5}}-5=32-5=27\ne RHS$
So, 5 is also not a solution.

Put x = 6,
$LHS={{2}^{6}}-6=64-6=58=RHS$

Hence, x = 6 is the solution of the given equation.

So, total numbers of candidates present are 6.

Note: One can prove the used equation

$=C_{0}^{n}+C_{1}^{n}+C_{2}^{n}+.............C_{n}^{n}={{2}^{n}}$
Let us write the expansion of ${{\left( 1+x \right)}^{n}}$ by binomial theorem. Hence, we get
${{\left( 1+x \right)}^{n}}=C_{0}^{n}+C_{1}^{n}x+C_{2}^{n}{{x}^{2}}+C_{3}^{n}{{x}^{3}}+.......+=C_{n}^{n}{{x}^{n}}$

Now, put x = 1 to both sides, we get
${{2}^{n}}=C_{0}^{n}+C_{1}^{n}+C_{2}^{n}+.............C_{n}^{n}$
Hence, proved
Equation ${{2}^{x}}-x=58$ can only be solved by hit-trial method. There is no other way to get the value of ‘x’ at this level.

Hence, don’t be confused by the equation.