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? A. 32
Answers
Answer:
From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that
at least one black ball should be there.
Hence we have 3 choices
All three are black
Two are black and one is non black
One is black and two are non black
Total number of ways
= 3C3 + (3C2 x 6C1) + (3C1 x 6C2) [because 6 are non black]
=1+[3×6]+[3×(6×52×1)]=1+18+45=64
Given : a box contains 2 white balls, 3 black balls and 4 red balls
To Find : the number of ways in which 3 balls can be drawn from the box so that at least one of the balls is black
Solution:
White = 2
Black = 3
Red = 4
Total = 2 + 3 + 4 = 9
3 Balls Can be drawn out of 9 balls in ⁹C₃ = 84 Ways
at least one of the balls is black = 120 - No black Balls
No black Balls means 3 balls out of remaining 6 balls = ⁶C₃ = 20 Ways
the number of ways in which 3 balls can be drawn from the box so that at least one of the balls is black = 84 - 20 = 64 Ways
other method :
1 black , 2 black , 3 black
³C₁*⁶C₂ + ³C₂.⁶C₁ + ³C₃.⁶C₀
= 3 * 15 + 3* 6 + 1 * 1
= 45 + 18 + 1
= 64
64 ways in which 3 balls can be drawn from the box so that at least one of the balls is black
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