No of selections and probability

Six balls are selected from a set of 44 balls, numbered from 1 to 44. The balls are colored. The balles numbered, 1, 5, 9 ... are red, 2,6,10 are orange, the balls no, 3 ,7, 11 are yellow and the balls numbered 4,8,12... are pink

What is the probability that the selected six balls contain:
f)Four colours:

Heres my attempt: there are two cases:
[1,1,1,3] and [1,1,2,2]

This equates to
(4C3 x (11C1)^3 x (11C3) + 4C2 x (11C1)^2 x (11C2)^2)/44C11

However the book has a different solution:
(4C3 x (11C1)^3 x (11C3) + 4C2 x 2C1 X (11C1)^2 x (11C2)^2)/44C11

Can someone explain to me?

Quote:

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Originally Posted by Lukybear

Six balls are selected from a set of 44 balls, numbered from 1 to 44. The balls are colored. The balles numbered, 1, 5, 9 ... are red, 2,6,10 are orange, the balls no, 3 ,7, 11 are yellow and the balls numbered 4,8,12... are pink

What is the probability that the selected six balls contain:
f)Four colours:

Heres my attempt: there are two cases:
[1,1,1,3] and [1,1,2,2]

This equates to
(4C3 x (11C1)^3 x (11C3) + 4C2 x (11C1)^2 x (11C2)^2)/44C11

However the book has a different solution:
(4C3 x (11C1)^3 x (11C3) + 4C2 x 2C1 X (11C1)^2 x (11C2)^2)/44C11

Can someone explain to me?

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shouldn't the denominator be 44C6?

the 2C1 is there because {1 red, 1 orange, 2 yellow, 2 pink} is different from {2 red, 2 orange, 1 yellow, 1 pink}.

Yea it should be 44C6. But isnt the 2C1 counted for by 4C2?

Quote:

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Originally Posted by Lukybear

Yea it should be 44C6. But isnt the 2C1 counted for by 4C2?

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This is under case {1,1,2,2}

4C2 means you're selecting two objects from four. Think about what exactly it is you are selecting. Say you select red and orange. Do you assign these to {1,1} or to {2,2}? It could be either. Therefore you multiply by 2 which equals 2C1.

Yes that makes more sense.Thanks

Hmm. After it made sense, it had seemed that the answer was wrong after all. There is no 4C2 :(. Can you help me make sense of it now.

Quote:

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Originally Posted by Lukybear

Hmm. After it made sense, it had seemed that the answer was wrong after all. There is no 4C2 :(. Can you help me make sense of it now.

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I am rather puzzled. I would like to write a brute force computer program (since 44C6 is pretty small) and compare results, then get to the bottom of it. I may or may not have time to do that right now.

Quote:

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Originally Posted by undefined

I am rather puzzled. I would like to write a brute force computer program (since 44C6 is pretty small) and compare results, then get to the bottom of it. I may or may not have time to do that right now.

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Oh all right, as often happens I have made a silly error. Sorry about that.

Your original answer is right, and the book's answer is wrong.

Once you select the two colours to correspond to (for example) {1,1} then the colours for {2,2} are fixed, and by multiplying by 2 you double count.

Here's the Java program I wrote

Code:

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import java.util.ArrayList;

public class ColouredBallsProb {
    static int c;
   
    public static void main(String[] args) {
        long t=time();
        c=0;
        recurse(0, new ArrayList<Integer>());
        System.out.println(c);
        System.out.println("elapsed: "+(time()-t)/1000.0+" s");
    }
   
    static void recurse(int ind, ArrayList<Integer> balls) {
        if(balls.size()==6) {
            if (oneOfEach(balls)) c++;
            return;
        }
        if(ind>43) return;
        // select or don't select current ball
        recurse(ind+1,balls);
        ArrayList<Integer> nextBalls=new ArrayList<Integer>(balls);
        nextBalls.add(ind);
        recurse(ind+1,nextBalls);
    }
   
    static boolean oneOfEach(ArrayList<Integer> balls) {
        // assume 6 balls
        int i;
        boolean[] colours=new boolean[4];
        for (i=0;i<6;i++) {
            colours[balls.get(i)%4]=true;
        }
        for (i=0;i<4;i++)
            if(!colours[i])
                return false;
        return true;
    }
   
    static long time() {
        return System.currentTimeMillis();
    }
}

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Output

Code:

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3074610
elapsed: 1.101 s

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As you can see it matches the first answer you gave (just the numerator). Sorry for the mix up.

Wow. That is brilliant. Absolutely fantastic.

I get your fixing explanation. But what about multiplying by 2? Wheres that?

Quote:

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Originally Posted by Lukybear

Wow. That is brilliant. Absolutely fantastic.

I get your fixing explanation. But what about multiplying by 2? Wheres that?

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I don't follow.. The whole issue is whether the 2C1 in bold in post #1 should appear in the answer right? It shouldn't.

When you wrote 4C2 in post #6 I assumed it was a typo and was supposed to be 2C1.

The answer is

(4C3 x (11C1)^3 x (11C3) + 4C2 x (11C1)^2 x (11C2)^2)/44C6

Hope it's clear now?

Yea thanks.