Question

In an election 3 persons are to be elected from 6 candidates. A voter can cast any
number of votes but not more than the candidates to be elected. In how many ways
Can he cast his vote?
$
  {\text{A}}{\text{. 41}} \\
  {\text{B}}{\text{. 20}} \\
  {\text{C}}{\text{. 15}} \\
  {\text{D}}{\text{. 6}} \\
$

Answer 1

Hint: In this type of question first we calculate the different number of ways in which a voter can cast his vote. Since, here different options are available to whom a voter cast his vote so for calculating the number of ways we have to use the Combination Concept of $_{}^n{C_r}$.

Complete step-by-step answer:
Given,
number of persons elected in an election=3
 Total number of candidates =6
And a voter can cast any number of votes but not more than number of candidates to be elected i.e. 3
Thus, a voter can cast a vote 1 or 2 or 3 times.
If he votes one time ,the number of ways he can cast his vote=$_{}^6{C_1}$ -- eq.1
If he votes two times ,the number of ways he can cast his vote=$_{}^6{C_2}$ ----eq.2
If he votes three times ,the number of ways he can cast his vote=$_{}^6{C_3}$ ---eq.3
Thus, total number of ways in which he can cast his vote =
$
   \Rightarrow {\text{ }}_{}^6{C_1} + _{}^6{C_2} + _{}^6{C_3} \\
   \Rightarrow {\text{ }}\dfrac{{6!}}{{1!(5!)}} + \dfrac{{6!}}{{2!(4!)}} + \dfrac{{6!}}{{3!(3!)}} \\
   \Rightarrow {\text{ }}6{\text{ }} + {\text{ 15 + 20}} \\
   \Rightarrow {\text{ 41 }} \\
 $
Hence, option A. is correct.

Note:-Whenever you get this type of problem the key concept of solving this is to make possible combinations of number of ways using formula $_{}^n{C_r}$. And then add them to get the answer. Here one more thing to be noted a voter must have to cast at least one vote so we do not add $_{}^6{C_0}$ to the total number of ways in which a voter can cast his vote .