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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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24??

Let the number of chosen green, yellow and red balls be g, y and 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 yellow balls (which of course is violating the upper limit), now we have to choose only 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 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:

$(i) – (ii) – (iii) + (iv)$

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

$=66-15-28+1$

$=24$

by (855 points)
edited
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Thanks bro for explanation

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 ago by (513 points)