# No of selections and probability

• Sep 26th 2010, 07:57 PM
Lukybear
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?
• Sep 26th 2010, 08:30 PM
undefined
Quote:

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?

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}.
• Sep 26th 2010, 09:36 PM
Lukybear
Yea it should be 44C6. But isnt the 2C1 counted for by 4C2?
• Sep 26th 2010, 09:59 PM
undefined
Quote:

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

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.
• Sep 27th 2010, 04:03 AM
Lukybear
Yes that makes more sense.Thanks
• Sep 27th 2010, 05:15 AM
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.
• Sep 27th 2010, 05:20 AM
undefined
Quote:

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.

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.
• Sep 27th 2010, 05:42 AM
undefined
Quote:

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.

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

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:

```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();     } }```
Output

Code:

```3074610 elapsed: 1.101 s```
As you can see it matches the first answer you gave (just the numerator). Sorry for the mix up.
• Sep 28th 2010, 01:05 AM
Lukybear
Wow. That is brilliant. Absolutely fantastic.

I get your fixing explanation. But what about multiplying by 2? Wheres that?
• Sep 28th 2010, 05:01 AM
undefined
Quote:

Originally Posted by Lukybear
Wow. That is brilliant. Absolutely fantastic.

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

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.