In an election,a voter can vote for any numbe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The number of candidates to be elected is $4$ out of $10$. Since a voter can vote for any number of candidates from $1$ to $4$,the total number of

Vedclass · Accepted Answer

$385$

Vedclass · Answer

(A) The number of candidates to be elected is $4$ out of $10$. Since a voter can vote for any number of candidates from $1$ to $4$,the total number of ways is given by the sum of combinations: Total ways $= ^{10}C_1 + ^{10}C_2 + ^{10}C_3 + ^{10}C_4$. Calculating each term: $^{10}C_1 = 10$ $^{10}C_2 = \frac{10 	imes 9}{2 	imes 1} = 45$ $^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$ $^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$ Total ways $= 10 + 45 + 120 + 210 = 385$.

In an election,a voter can vote for any number of candidates but not more than the number of candidates to be elected. There are $10$ candidates and $4$ are to be elected. If a voter must vote for at least one candidate,in how many ways can they vote?

Similar Questions

If $20$ identical books are to be distributed among $4$ persons such that each person receives at least one book,in how many ways can this be done?

The number of ways to select $n$ objects from $2n$ objects, of which $n$ are identical and the rest are different, is:

The number of $5$ card combinations out of a deck of $52$ cards such that there is exactly one ace in each combination is:

If $\binom{n-1}{4}, \binom{n-1}{5}$ and $\binom{n-1}{6}$ are in Arithmetic Progression,then find the relation.

$A$ student is to answer $10$ out of $13$ questions in an examination such that he must choose at least $4$ from the first $5$ questions. The number of choices available to him is

Vedclass Test Series

Exam Paper Generator

Online Exam Module