There are 4 red, 3 green and 6 white balls in a jar. In how many ways is it possible to choose 6 balls if there must be one of each color?

Question

There are 4 red, 3 green and 6 white balls in a jar. In how many ways is it possible to choose 6 balls if there must be one of each color?

I think I'm supposed to use the binomial coefficient but I can't figure out how to use it in examples like these.

Answer 1

Hint: One way to do this is to find the total number of ways you can choose $6$ balls (regardless of color), then subtract the following three quantities:

• The number of ways to choose $6$ balls, all red or green
• The number of ways to choose $6$ balls, all red or white
• The number of ways to choose $6$ balls, all white or green

Now notice that you've double-counted, so you have to add back, the following three quantities:

• The number of ways to choose $6$ balls, all red
• The number of ways to choose $6$ balls, all white
• The number of ways to choose $6$ balls, all green

Answer 2

The total number of ways in which all the balls can be selected is $(13C6) = \frac{13!}{6!(13-6)!}$

The invalid cases are :

• All $6$ are balls white and red balls $(10C6)$

• All $6$ balls are white and green balls $(9C6)$

• All $6$ balls being white is counted twice

• All $6$ balls are red and green balls $(7C6)$

Therefore, valid combinations are ${(13C6) + 1 - ( (10C6) + (9C6) + (7C6) )} = 1716 + 1 -(210 + 84 + 7) =1415+1 = 1416$

Answer 3

Assume for the moment that we have an unlimited number of balls of each color. The told story then amounts to arranging six uncolored balls in a row, then putting a separator into two of the five spaces between two adjacent balls, and finally painting the balls of the first emerging group red, the balls of the second group green, and the balls of the third group white. There are ${5\choose2}=10$ ways to put the separators. Out of these ten ways exactly one is forbidden, namely coloring $4$ balls green and the remaining two red and white.

It follows that there are $9$ different kinds of drawings.

Answer 4

There are $9$ possible outcomes, and no matter identical or not, any two balls of the same color are indeed different.

Combination with repetition: 4, 3, 6
r, g, w:
4, 3, 6
OOO||: $(1,1,1)+(3,0,0)={4\choose4}\cdot{3\choose1}\cdot{6\choose1}=18.$
OO|O|: $(1,1,1)+(2,1,0)={4\choose3}\cdot{3\choose2}\cdot{6\choose1}=72.$
OO||O: $(1,1,1)+(2,0,1)={4\choose3}\cdot{3\choose1}\cdot{6\choose2}=12\cdot15=180.$
O|OO|: $(1,1,1)+(1,2,0)={4\choose2}\cdot{3\choose3}\cdot{6\choose1}=36.$
O||OO: $(1,1,1)+(1,0,2)={4\choose2}\cdot{3\choose1}\cdot{6\choose3}=18\cdot20=360.$
O|O|O: $(1,1,1)+(1,1,1)={4\choose2}\cdot{3\choose2}\cdot{6\choose2}=18\cdot15=270.$
|OO|O: $(1,1,1)+(0,2,1)={4\choose1}\cdot{3\choose3}\cdot{6\choose2}=60.$
|O|OO: $(1,1,1)+(0,1,2)={4\choose1}\cdot{3\choose2}\cdot{6\choose3}=12\cdot20=240.$
||OOO: $(1,1,1)+(0,0,3)={4\choose1}\cdot{3\choose1}\cdot{6\choose4}=12\cdot15=180.$

The total $= 1416.$