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How many ways can 10 balls be chosen from a container having 10 identical green balls , 5 identical yellow balls and 3 identical blue balls

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Let the number of chosen green, yellow and red balls be **g****, y **and **r **respectively.

Now, the limits on these variables will be $0\leq g \leq 10, 0\leq y \leq 5, \text{ and } 0\leq r \leq 3.$

And we want to ensure $g+y+r=10$.

Let’s ignore the upper limits for now.

By using stars and bars (which says that $x_1+x_2+\dots+x_k=n$ has total $^{n+k-1}C_{k-1}$ solutions under the constraint that $\forall i \in [1,k], x_i$ is a non-negative Integer), we can say that the number of solutions for the above equation is:

$^{10+3-1}C_{3-1}=\ ^{12}C_2 \ \ \ \ \ \dots \dots (i)$

Since we ignored the upper limits, we have also counted the solutions that violate our constraints, now we need to subtract the counts of such solutions.

**Number of solutions violating upper limit for g:**

0, because over all sum is **10, **so it is guaranteed that any value of **g >10 **is not counted.

**Number of solutions violating upper limit for y:**

Let us assume that we have already chosen **6 **yellow balls (which of course is violating the upper limit), now we have to choose only **4 **balls, so the number of solutions in this case will be:

$^{4+3-1}C_{3-1}=\ ^6C_2 \ \ \ \ \ \dots \dots (ii)$

**Number of solutions violating upper limit for r:**

Let us assume that we have already chosen **4**** **red balls, now we have to choose only **6**** **balls, so the number of solutions in this case will be:

$^{6+3-1}C_{3-1}=\ ^8C_2 \ \ \ \ \ \dots \dots (iii)$

If we subtract **(ii) **and **(iii) **from **(i), **then we will end up subtracting the number of solutions in which the constraints for** **both **y **and **r **were violated twice, which we were supposed to subtract only once, so we need to add this value back to our answer.

**Number of solutions violating upper limits for y as well as r:**

Let us assume that we have already chosen **6 **yellow and **4**** **red balls, now we have to choose **0**** **balls, so the number of solutions in this case will be:

$^{0+3-1}C_{3-1}=\ ^2C_2 \ \ \ \ \ \dots \dots (iv)$

So our final solution is:

$(i) – (ii) – (iii) + (iv)$

$=\ ^{12}C_2 –\ ^6C_2\ – \ ^8C_2 +\ ^2C_2$

$=66-15-28+1$

$=24$

0 votes

It can be solved by generating function

I am taking different values of power of x from green and taking remaining power from product of yellow and blue

Let us chose x^10 from Green

then we will have 1 from Yellow and BLue – only 1 way

if we chose x^9 from green

we will require coefficient of x from multiplication of second and third term – 2 ways

and so on calculate

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