A box contains 5 blue, 6 yellow, and 4 red balls. The number of ways, of drawing 8 balls containing atleast two balls of each colour, is :

Let $$b,y,r$$ denote the numbers of blue, yellow and red balls drawn, respectively.

Required conditions:
  • $$b+y+r = 8$$ (total balls drawn)
  • $$b \ge 2,\; y \ge 2,\; r \ge 2$$ (at least two of each colour)
  • Availability limits: $$b \le 5,\; y \le 6,\; r \le 4$$

We first list all integral triples $$(b,y,r)$$ satisfying these inequalities.

Case 1: $$b = 2 \;\Rightarrow\; y+r = 6$$

  Possible pairs $$\Rightarrow (y,r) = (2,4),(3,3),(4,2)$$ — three triples.

  Thus $$\;(2,2,4),(2,3,3),(2,4,2)$$.

Case 2: $$b = 3 \;\Rightarrow\; y+r = 5$$

  Possible pairs $$\Rightarrow (y,r) = (2,3),(3,2)$$ — two triples.

  Thus $$\;(3,2,3),(3,3,2)$$.

Case 3: $$b = 4 \;\Rightarrow\; y+r = 4$$

  Only pair satisfying $$y,r \ge 2$$ is $$(2,2)$$.

  Thus $$\;(4,2,2)$$.

No solution exists for $$b = 5$$ because it would give $$y+r = 3 \lt 4$$, contradicting $$y,r \ge 2$$.

Hence the feasible colour distributions are:
$$(2,2,4),\;(2,3,3),\;(2,4,2),\;(3,2,3),\;(3,3,2),\;(4,2,2).$$

For each triple, the balls of a given colour are distinct, so the number of selections is the product of binomial coefficients:

$$ \begin{aligned} (2,2,4):\;&\binom{5}{2}\binom{6}{2}\binom{4}{4} &= 10 \times 15 \times 1 &= 150\\ (2,3,3):\;&\binom{5}{2}\binom{6}{3}\binom{4}{3} &= 10 \times 20 \times 4 &= 800\\ (2,4,2):\;&\binom{5}{2}\binom{6}{4}\binom{4}{2} &= 10 \times 15 \times 6 &= 900\\ (3,2,3):\;&\binom{5}{3}\binom{6}{2}\binom{4}{3} &= 10 \times 15 \times 4 &= 600\\ (3,3,2):\;&\binom{5}{3}\binom{6}{3}\binom{4}{2} &= 10 \times 20 \times 6 &= 1200\\ (4,2,2):\;&\binom{5}{4}\binom{6}{2}\binom{4}{2} &= 5 \times 15 \times 6 &= 450 \end{aligned} $$

Adding all these cases:
$$150 + 800 + 900 + 600 + 1200 + 450 = 4100.$$

Therefore, the number of ways to draw the 8 balls is $$4100$$.

Option A which is: $$4100$$