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A jar holds 4 red, 3 green and 6 white marbles. How many different ways can you pick 6 marbles so that you have at least one of each color"
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If 1 red , 1 green and 1 white are chosen, the we need to choose 3 more marbles from out of 3 red, 2 green, and 6 white marbles. So, Form 3+2+5=10 marbles we need to choose 3 marbles. This can be done \(^{10}C\)\(^3\)Ways. \(={10*9*8*7!}/{7!*3!}\) =120
A jar holds 4 red, 3 green and 6 white marbles. How many different ways can you pick 6 marbles so that you have at least one of each color"
sorry : The source did not have any answer choices.
If 1 red , 1 green and 1 white are chosen, the we need to choose 3 more marbles from out of 3 red, 2 green, and 6 white marbles. So, Form 3+2+5=10 marbles we need to choose 3 marbles. This can be done \(^{10}C\)\(^3\)Ways. \(={10*9*8*7!}/{7!*3!}\) =120
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Far too many possibilities. The balls of the same color are assumed to be identical, what matters is just the number of balls of each color. R - red, G - green, W - white After we have chosen one of each type, we need 3 more balls to chose from 3R, 2G and 5W. The question is about choosing the colors rather than specific balls.
We can again choose one of each color - 1R, 1G, 1W - 1 possibility 2 of one color and the third one of a different color - choose the color for two balls, then the color for the one ball - 3*2 = 6 possibilities 3 balls of the same color - only 2 possibilities - we don't have 3G balls left
Therefore, a total of 1+6+2=9 possibilities.
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