probability question 2 (1 Viewer)

[mathsbrain]

Question: How many ways for 5 DIFFERENT marbles to go in 3 jars so that 2 jars contains 2 marbles and the last jar contains one marble?

Solution: 5C2 * 3C2 * 1C1 * 3! / 2!

What my problem is: i understand that we firstly choose 2 out of 5, then 2 out of the remaining 3, then 1 out of 1, then 3! is for the different arrangements. It's the divide by 2! thats the problem. The jars and marbles are all different, why are we dividing by 2! ? any similar examples to illuminate this idea?

Also, say when you choose 2 boys from 3 boys, 3C2=3. One can try to do 3C1 * 2C1=6 and realise they have to divide by 2!(by the way when do we know to do this? Is it when we choosing from the same category?) because of ordering problem. But how come here 5C2 *3C2 *1C1 we don't have to? It seems only logical to divide by 3! here???

 

[townie]

I *think* you divide by 2! Because the order of the first 2 jars doesn't matter (whereas the third jar has to be the third jar) but I haven't done probability in many years so I could be wrong

 

[math man]

simple, first we choose 2 marbles to go into one jar, 5C2, next we choose 2 of remaining 3, 3C2, to go into 2nd jar and we have one left for 5th jar. Now
we treat the jars with 2 marbles each as identical so to arrange the current marbles across the jars it is 3!/2!

 

[mathsbrain]

simple, first we choose 2 marbles to go into one jar, 5C2, next we choose 2 of remaining 3, 3C2, to go into 2nd jar and we have one left for 5th jar. Now
we treat the jars with 2 marbles each as identical so to arrange the current marbles across the jars it is 3!/2!

I still don't get why the jars with 2 marbles are identical? aren't the jars different and so are the marbles?

 

[D94]

I still don't get why the jars with 2 marbles are identical? aren't the jars different and so are the marbles?

What makes one jar more superior than the other jar? We can't differentiate between choosing one jar over another, so they are essentially identical. We know the 3rd jar can't be treated as identical because we explicitly called it the "last jar", which implies we've already used 2, so that last jar is unique.