Combination (Repetition)

Question

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?

Comments

10 ??

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Yes, I am also getting 10 but I don't know its generalize formula?

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no generalized formula needed,
first take case of only 2 same color balls out of 4 selected balls
secondly 3 same colors balls
finaly all same colors balls.

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yes, it will be 13..

i had not considerd 2 balls of same color and 2 balls of another same color..

Answer 1

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.

Comments

These are the 13 possibilities:

R  G  B

3  1  0

3  0  1

1  3  0

0  3  1

1  0  3

0  1  3

2  1  1

1  2  1

1  1  2

2  2  0

0  2  2

2  0  2

0  0  4

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but when I am talking number of balls to be picked as 5 then correct answer should be 14 but, through your approach it is giving as 16 as:

C(5 + 3 - 1, 3 - 1) = 21

removing those cases when 5 red, 4 red, 5 green, 4 green and  5 blue balls are picked.

21 - 5 = 16

which is not 14.

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The number of cases to be removed is 7 (not 5).

R G B

5 0 0

4 1 0

4 0 1

0 5 0

1 4 0

0 4 1

0 0 5

Hope that helps :)

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Thanks!!!

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when the difference is high then how can we do that?

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then it could be so hard ?

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Suppose we want to choose 7 8 balls from the given set of balls?

Answer 2

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)² (1-x^5)(1-x)^-3      

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

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

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

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  :)