A box contains 2 white balls,3 black balls an | vedclass

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

Question

(C) Total number of balls $= 2 + 3 + 4 = 9$.Number of black balls $= 3$.Number of non-black balls $= 2 + 4 = 6$.We need to select $3$ balls such that

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(C) Total number of balls $= 2 + 3 + 4 = 9$.Number of black balls $= 3$.Number of non-black balls $= 2 + 4 = 6$.We need to select $3$ balls such that at least one is black.This can be done in the following ways:$1$ black and $2$ non-black balls: ${}^{3}C_{1} 	imes {}^{6}C_{2} = 3 	imes 15 = 45$.$2$ black and $1$ non-black balls: ${}^{3}C_{2} 	imes {}^{6}C_{1} = 3 	imes 6 = 18$.$3$ black and $0$ non-black balls: ${}^{3}C_{3} 	imes {}^{6}C_{0} = 1 	imes 1 = 1$.Total ways $= 45 + 18 + 1 = 64$.

$A$ box contains $2$ white balls,$3$ black balls and $4$ red balls. In how many ways can $3$ balls be drawn from the box,if at least one black ball is to be included in the draw?

Similar Questions

In how many ways can a football team of $11$ players be selected from $15$ players? In how many cases can a particular player be included?

How many different four-letter words can be formed (the words need not be meaningful) using the letters of the word $MEDITERRANEAN$ such that the first letter is $E$ and the last letter is $R$?

$P_1$ and $P_2$ are two distinct and intersecting planes. Three non-collinear points lie on $P_1$ and another three non-collinear points lie on $P_2$ (none being on the line of intersection of the planes). Then,the maximum number of tetrahedrons formed using these six points is:

The product of any $r$ consecutive natural numbers is always divisible by

The sum of all $4$ digit numbers that can be formed by using the digits $2, 4, 6, 8$ (repetition of digits not allowed) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module