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A girl has to choose 4 items from a bucket which contains 3 red, 3 green, and 4 blue balls, now in how many ways she can do this?

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For now, let's consider that there are an infinite number of balls of each colour. So, we have to find the number of ways the girl can choose a total of 4 balls.

Think of colours as boxes, and we have 4 items to place in 3 different boxes. Visualise items as stars and boxes by the separators

One combination is:  * | * | * *  (1 red, 1 green, and 2 blue balls)

or:  * | | * * *  (1 red, 0 green, and 3 blue balls)

So, now we have to arrange 4 stars in 4 + 3 - 1 places (3 - 1 are the separators)

i.e. C(4 + 3 - 1, 4) = C(6, 4) = 15

Now, remove those cases when she chooses 4 red or 4 green balls.

So, the answer should be 15 - 2 = 13.
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Another way This problem is similar as chocolate in box problem 

(x^0+x^1+x^2+x^3)(x^0+x^1+x^2+x^3)(x^0+x^1+x^2+x^3+x^4)

(1-x^4)(1-x^5)(1-x)-3      

Now solve it further (1+x^8-2x^4)(1-x^5)$\binom{2+r}{r}$xr

now multiply those term in power of x comes out to be x^4 because we need 4 item  

(1-2x^4)$\binom{2+r}{r}$xr

now at x=4 bcz multiply x^r with 1  

$\binom{2+4}{4}$x^4  -2x^4 now calculate the coffecient of power of x^4

$\binom{6}{4}$  -2

15-2

13  :)

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