A box contains 3, 2, and 4 balls of red, yellow, and green, respectively. What is the number of ways to arrange 6 chosen balls in order?
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A box contains 3, 2, and 4 balls of red, yellow, and green, respectively. What is the number of ways to arrange 6
Answer:
The bag contains 2 White, 3 Black and 4 Red balls.
So, total 9 balls are there in the bag; among them 3 are Black and 6 are non-Black balls.
Three balls can randomly be drawn in (9C3) = 84 ways.
1 Black and 2 non-Black balls can be drawn in (3C1)*(6C2) = 45 ways.
1 non-Black and 2 Black balls can be drawn in (6C1)*(3C2) = 18 ways.
3 Black balls can be drawn in (3C3) = 1 way.
So, three balls drawn in (45 + 18 + 1) = 64 ways will have at least one Black ball among the drawn ones.
So, the probability of getting at least one Black ball among the three drawn =
(64 / 84) = (16 / 21).
Verified answer
Before we start, let me tell you one thing. Using the ordinary technique of permutation and combinations, the solution will become more cumbersome. We should use some another method for clarity. I will use exponential generating function this time. Have a look at article on wikipedia about E.G.F. first so that you can understand it better way.
So first of all, the order in the question refers to the order of the colour like RRRYGG or RRYYGG etc. where R, Y and G denotes Red, Yellow and Green respectively.
EGF for balls coloured red:
[tex]\displaystyle\left(1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!} \right)[/tex]
EGF for balls coloured yellow:
[tex]\displaystyle\left(1+\frac{x^1}{1!}+\frac{x^2}{2!}\right)[/tex]
EGF for balls coloured green:
[tex]\displaystyle\bigg(1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\bigg)[/tex]
Multiplying all the above cases and finding the coefficient of x^6/6!.
[tex]\displaystyle\bigg[\bigg(1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!} \bigg)\bigg(1+\frac{x^1}{1!}+\frac{x^2}{2!}\bigg)\bigg(1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\bigg)\bigg][/tex]
[tex]\displaystyle = \frac{x^{9}}{288}+\frac{x^{8}}{32}+\frac{23x^{7}}{144}+\frac{41x^{6}}{72}+\frac{3x^{5}}{2}+\frac{71x^{4}}{24}+\frac{13x^{3}}{3}+\frac{9x^{2}}{2}+3x+1[/tex]
Coefficient of x^6 = 41/72.
So the coefficient of x^6/6! = 41/72 * 6! = 410.
So the required number of ways is 410.