Question

Members to be elected. If a voter can vote in 30 different ways, find the number of candidates standing for the election if every voter has to vote for at least one candidate and not more than the number of candidates to be elected​

Answer 1

Given :- Members to be elected. If a voter can vote in 30 different ways, find the number of candidates standing for the election if every voter has to vote for at least one candidate and not more than the number of candidates to be elected ?

Answer :-

Let us assume that, the number of candidates standing for the election are n .

so,

→ n(C)1 + n(C)2 + n(C)3 + ____ n(C)(n - 1) = 32

→ 2^n - 2 = 30

→ 2^n = 30 + 2

→ 2^n = 32

→ 2^n = 2⁵

→ n = 5

therefore, there were 5 candidates standing for the election .

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Answer 2

Given : Members to be elected.

A voter can vote in 30 different ways,

To Find : the number of candidates standing for the election if every voter has to vote for at least one candidate and not more than the number of candidates to be elected​

Solution:

number of  candidates standing for the election = n

number of candidates to be elected​ = x

x  ≤  n

ⁿC₁  + ⁿC₂  +  ______  + ⁿCₓ   = 30

case 1 :  x = 1

=>  ⁿC₁ = 30

Hence 30  candidates

case 2 :

x = 2

=> ⁿC₁  + ⁿC₂  = 30

=> n  + (n ) (n - 1)/2  = 30

=> n²  + n -  60 = 0

no integral solution

case 3 :

x = 3

=> ⁿC₁  + ⁿC₂ +  ⁿC₃  = 30

=> n  + (n ) (n - 1)/2   +   (n ) (n - 1)(n-2)/3  = 30

no integral solution for n

case 4 :

x = 4

=> ⁿC₁  + ⁿC₂ +  ⁿC₃  + ⁿC₄ = 30

=> n = 5  

Hence 5 candidates

for x = 4  n  is 5    

Hence  no further    solution possible  as for x = 5 , even n = 5 will given 31

and this value will keep increasing with  x and n values

so Two possible solutions are

    30 candidates and 1 to be  elected

or  5 candidates  and 4 to be elected

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