A box contains 5 blue, 6 yellow and 4 red balls.

Question

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

Accepted Answer

Let us first understand the problem carefully. We are given a box containing:

Total number of blue balls = $5$

Total number of yellow balls = $6$

Total number of red balls = $4$

We need to draw a total of $8$ balls such that there are at least two balls of each colour. Let the number of blue, yellow, and red balls drawn be denoted by $b$, $y$, and $r$ respectively.

According to the question, we have the following equation:

$b+y+r=8$

Since we need at least $2$ balls of each colour, we must satisfy the conditions:

$b\ge 2,y\ge 2,r\ge 2$

Since $2+2+2=6$, we only need to decide how to distribute the remaining $2$ balls among the three colours. Let us list all the possible combinations for $(b,y,r)$:

Case 1: $4$ blue, $2$ yellow, $2$ red ($b=4,y=2,r=2$)

Case 2: $2$ blue, $4$ yellow, $2$ red ($b=2,y=4,r=2$)

Case 3: $2$ blue, $2$ yellow, $4$ red ($b=2,y=2,r=4$)

Case 4: $3$ blue, $3$ yellow, $2$ red ($b=3,y=3,r=2$)

Case 5: $3$ blue, $2$ yellow, $3$ red ($b=3,y=2,r=3$)

Case 6: $2$ blue, $3$ yellow, $3$ red ($b=2,y=3,r=3$)

Now, we will find the number of ways for each case using the combinations formula ${}^{n}{C}_{r}=\frac{n!}{r!(n-r)!}$.

1. Number of ways for Case 1:

${}^{5}{C}_{4}{	imes }^{6}{C}_{2}{	imes }^{4}{C}_{2}=5	imes 15	imes 6=450$

2. Number of ways for Case 2:

${}^{5}{C}_{2}{	imes }^{6}{C}_{4}{	imes }^{4}{C}_{2}=10	imes 15	imes 6=900$

3. Number of ways for Case 3:

${}^{5}{C}_{2}{	imes }^{6}{C}_{2}{	imes }^{4}{C}_{4}=10	imes 15	imes 1=150$

4. Number of ways for Case 4:

${}^{5}{C}_{3}{	imes }^{6}{C}_{3}{	imes }^{4}{C}_{2}=10	imes 20	imes 6=1200$

5. Number of ways for Case 5:

${}^{5}{C}_{3}{	imes }^{6}{C}_{2}{	imes }^{4}{C}_{3}=10	imes 15	imes 4=600$

6. Number of ways for Case 6:

${}^{5}{C}_{2}{	imes }^{6}{C}_{3}{	imes }^{4}{C}_{3}=10	imes 20	imes 4=800$

To find the total number of ways to draw the $8$ balls satisfying the given conditions, we simply add the number of ways from all the above cases:

$=450+900+150+1200+600+800$

$=4100$

Therefore, the total number of ways of drawing $8$ balls containing at least two balls of each colour is $4100$.

Option A is the correct answer.

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