
In an election there are 8 candidates, out of which 5 are to be chosen. If a voter may vote for any number of candidates but not greater than the number to be chosen, then in how many ways can a voter vote?
A. 216
B. 114
C. 218
D. None of these
Answer
164.1k+ views
Hint: A voter can vote only one person or 2 persons or 3 persons or 4 persons or 5 persons out of 8. We have to find the number ways to vote only one person or 2 persons or 3 persons or 4 persons or 5 persons out of 8 by using commination formula. Then add the ways to get required solution.
Formula Used:Combination formula:
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Complete step by step solution:There are 8 candidates in an election. A voter can vote for only 5 candidates. But a voter can vote any number of candidates which less than or equal to 5.
There are 5 cases to vote. A voter can vote one person or 2 persons or 3 persons or 4 persons or 5 persons out of 8.
Case I: Vote for 1 candidate
The number of ways to vote one candidate is
\[{}^8{C_1}\]
\[ = \dfrac{{8!}}{{1!\left( {8 - 1} \right)!}}\]
\[ = \dfrac{{8 \times 7!}}{{1!7!}}\]
\[ = 8\]
Case II: Vote for 2 candidates
The number of ways to vote two candidates is
\[{}^8{C_2}\]
\[ = \dfrac{{8!}}{{2!\left( {8 - 2} \right)!}}\]
\[ = \dfrac{{8 \times 7 \times 6!}}{{2!6!}}\]
\[ = 28\]
Case III: Vote for 3 candidates
The number of ways to vote three candidates is
\[{}^8{C_3}\]
\[ = \dfrac{{8!}}{{3!\left( {8 - 3} \right)!}}\]
\[ = \dfrac{{8 \times 7 \times 6 \times 5!}}{{3!5!}}\]
\[ = 56\]
Case IV: Vote for 4 candidates
The number of ways to vote four candidates is
\[{}^8{C_4}\]
\[ = \dfrac{{8!}}{{4!\left( {8 - 4} \right)!}}\]
\[ = \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4!4!}}\]
\[ = 70\]
Case V: Vote for 5 candidates
The number of ways to vote five candidates is
\[{}^8{C_5}\]
\[ = \dfrac{{8!}}{{5!\left( {8 - 3} \right)!}}\]
\[ = \dfrac{{8 \times 7 \times 6 \times 5!}}{{5!3!}}\]
\[ = 56\]
Now we will add all ways to find the required answer.
The total number of ways that a voter can vote is 8 + 28 + 56 + 70 + 56 = 218.
Option ‘C’ is correct
Note: Students often make mistake to solve such type of problems. They used permutation formula to solve it. As the order of selection does not matter, thus we will apply combination formula.
Formula Used:Combination formula:
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Complete step by step solution:There are 8 candidates in an election. A voter can vote for only 5 candidates. But a voter can vote any number of candidates which less than or equal to 5.
There are 5 cases to vote. A voter can vote one person or 2 persons or 3 persons or 4 persons or 5 persons out of 8.
Case I: Vote for 1 candidate
The number of ways to vote one candidate is
\[{}^8{C_1}\]
\[ = \dfrac{{8!}}{{1!\left( {8 - 1} \right)!}}\]
\[ = \dfrac{{8 \times 7!}}{{1!7!}}\]
\[ = 8\]
Case II: Vote for 2 candidates
The number of ways to vote two candidates is
\[{}^8{C_2}\]
\[ = \dfrac{{8!}}{{2!\left( {8 - 2} \right)!}}\]
\[ = \dfrac{{8 \times 7 \times 6!}}{{2!6!}}\]
\[ = 28\]
Case III: Vote for 3 candidates
The number of ways to vote three candidates is
\[{}^8{C_3}\]
\[ = \dfrac{{8!}}{{3!\left( {8 - 3} \right)!}}\]
\[ = \dfrac{{8 \times 7 \times 6 \times 5!}}{{3!5!}}\]
\[ = 56\]
Case IV: Vote for 4 candidates
The number of ways to vote four candidates is
\[{}^8{C_4}\]
\[ = \dfrac{{8!}}{{4!\left( {8 - 4} \right)!}}\]
\[ = \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4!4!}}\]
\[ = 70\]
Case V: Vote for 5 candidates
The number of ways to vote five candidates is
\[{}^8{C_5}\]
\[ = \dfrac{{8!}}{{5!\left( {8 - 3} \right)!}}\]
\[ = \dfrac{{8 \times 7 \times 6 \times 5!}}{{5!3!}}\]
\[ = 56\]
Now we will add all ways to find the required answer.
The total number of ways that a voter can vote is 8 + 28 + 56 + 70 + 56 = 218.
Option ‘C’ is correct
Note: Students often make mistake to solve such type of problems. They used permutation formula to solve it. As the order of selection does not matter, thus we will apply combination formula.
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