Question

A box contained \[2\] white balls, \[3\] black and \[4\] red balls. In how many ways can three balls be drawn from the box if atleast one black ball is to be included in the draw (the balls of the same colour are different).
A) \[60\]
B) \[64\]
C) \[56\]
D) None of the above

Answer 1

Hint: To find the number of ways three balls (colour independent) out of all the balls present we use the method of combination from permutation and combination and then after that we find the number of ways three balls out of six balls (non-black in colour) again with combination from permutation and combination and after that we subtract the number of ways three non-black balls from the number of ways of three colour independent balls. The combination formula used is:
\[{}^{N}{{C}_{R}},\text{ }{}^{M}{{C}_{R}}\]
where \[N,M\] are the total number of balls with black and without black balls respectively and \[R\] are the number of balls selected.

Complete step-by-step answer:
Placing the values given in the question in the formula we get:
The number of ways three balls are selected out of all the balls (included black balls) is equal to
\[{}^{N}{{C}_{R}}={}^{9}{{C}_{3}}\]
\[={}^{9}{{C}_{3}}\]
\[=\dfrac{9\times 8\times 7}{3\times 2\times 1}\]
\[=84\]
Now the number of ways three balls are selected out of all the balls (not including black balls) is equal to
\[{}^{M}{{C}_{R}}={}^{6}{{C}_{3}}\]
\[={}^{6}{{C}_{3}}\]
\[=\dfrac{6\times 5\times 4}{3\times 2\times 1}\]
\[=20\]
Hence, the number of ways three balls can be drawn from the box if atleast one black ball is to be included in the draw is \[84-20=64\].

Note: Another method to find the number of ways of selection of at least one black ball can be given in three ways:
Choice 1: One black and two colours. Using the Combination method we select one black from three black and two colors from six coloured balls.
\[{}^{3}{{C}_{1}}\times {}^{6}{{C}_{2}}\]
Choice 2: Two black and one colour. Using the Combination method we select two blacks from three black and one color from six coloured balls.
\[{}^{3}{{C}_{2}}\times {}^{6}{{C}_{1}}\]
Choice 3: Three black colours. Using the Combination method we select all three blacks from three black.
\[{}^{3}{{C}_{3}}\]
Hence, the number of ways can three balls be drawn from the box if atleast one black balls is to be included in draw is
\[{}^{3}{{C}_{1}}\times {}^{6}{{C}_{2}}+{}^{3}{{C}_{2}}\times {}^{6}{{C}_{1}}+{}^{3}{{C}_{3}}\]
\[=3\times \dfrac{6\times 5}{3\times 2}+\dfrac{3\times 2}{2\times 1}\times 6+1\]
\[=64\].