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A box contains 6 balls which may be all of different colors or three each of two colors or two each of three different colors. The number of ways of selecting 3 balls from the box (if a ball of the same color is identical) is N, then find the value of ‘N’.

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Last updated date: 17th Jun 2024
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Answer
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Hint: First we find out the number of ways the distribution can be done in all the 3 cases. The cases being 6 distinct colored balls or 6 balls three each of two colors or 6 balls of two each of three colors. We find the number of ways each case can be done and add them to find the final answer.

Complete step-by-step solution
A box contains 6 balls which may be all of different colors or three each of two colors or two each of three different colors.
So, the number of cases of possible options is 3.
Case 1: 6 distinct colored balls
We need to choose 3 balls out of 6 distinct ones which will be done in ${}^{6}{{C}_{3}}=\dfrac{6!}{3!\times 3!}=20$.
Case 2: 6 balls three each of two colors
We need to choose 3 balls. The choices can be 2 same-colored balls and 1 different or all three of same-colored balls.
The first part (2 white 1 black or 2 black 1 white) can be done in
$2\left[ {}^{3}{{C}_{2}}\times {}^{3}{{C}_{1}} \right]=2\times \dfrac{3!}{2!\times 1!}\times \dfrac{3!}{2!\times 1!}=18$.
The second part (3 white or 3 black) can be done in ${}^{3}{{C}_{3}}+{}^{3}{{C}_{3}}=1+1=2$.
Case 3: 6 balls of two each of three colors
We need to choose 3 balls. The choices can be 2 same-colored balls and 1 different or all three of different colored balls.
The first part can be done in choosing which color we are taking 2 balls off.
\[{}^{3}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}=3\times 2\times 2=12\].
The second part (1 white and 1 black and 1 red) can be done in \[{}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}={{2}^{3}}=8\].
Total number of options are $20+18+2+12+8=60$. The value of N is 20.

Note: We need to separately find the ways of choosing and we can use a table and actual color names to solve the problem. The multiplication of choosing is happening as the events are independent of each other.