What is the best way to solve this problem?

There is a bin with 12 objects labelled A, B and C. In the bin, there are 3 identical A objects; 4 identical B objects, and 5 identical C objects.

Someone takes 4 objects, at random, from the bin. What is the probability that this person took at least one object of each type? In other words, if he takes 4 objects, what is the probability that he took at least one object labelled A, one object labelled B, and one object labelled C.

Try and give the most efficient method to solve this problem you can think of.

(3*4*5*9 / 2!) / C(12,4) = 6/11

3*4*5 is the # ways to pick one of each.

Then once that's achieved, there are 9 remaining objects, one of which will be the extra. However, multiplying 60*9 double-counts because for the repeated symbol you've essentially used a permutation instead of a combination, thus counted order where there is none. So we divide out the 2! orders to make it a combination (since I'm using combos in the denominator).

That's 270 combos out of a possible C(12,4).

If you understand all that, try it using permutations instead of combos.

You can also see it this way;

1- P(at least one missing)

P(at least one missing)=P(A missing)+P(B missing)+P(C missing)-P(A and B missing)- P(A and C missing) -P( B and C missing)

Notice P (B and C missing ) =0 here (not possible with 4 chosen and A having only 3 ).

so P(at least one missing)= ((4+5,4)+(3+5,4)+(3+4,4)-(5,4)-(4,4))/(12,4)=5/11

So the required is 1-P(at least one missing)=6/11


notice; (m,n) means choose n from m or m!/(n!*(m-n)!)

I understand your creative explanation completely except for 1 part. My misunderstanding involves the formula: P(at least one missing) = P( A missing) +P(B missing) + P(C missing)-P(A and B missing)- P(A abd C missing)- P(B and C missing)? Why isn't the formula expressed as: P(at least one missing)=P(A missing)+P(B missing)+P(C missing)+P(A and B missing)+ P(A and C missing) +P( B and C missing)?
Thank you!

heehaw, why does multiplying 60*9 performs a permutation instead of a combination? And why does dividing by 2! transforms the permutation into combinations? Basically, I don't see how we counted order ... Thank you!

Because its like;



(seen as A missing or B missing or C missing, last term missing because its impossible to happen)
For example see this

https://en.wikipedia.org/wiki/Inclus...sion_principle

Also in terms of sets;




You correct for over counting basically because eg in P( A missing)+ P(B missing) we count twice P(A and B missing) so it has to be corrected by subtracting it one time.

masque de Z, got it! Makes sense!

I think it performs a permutation because the repeated symbol/object always comes in last. However, why does dividing by 2! transform the permutation into a combination?

This is the same principle as the original definition of permutation and combination though it may be a little harder to see.

Think of 8 identical items. When you say that there are 8*7 ways to select two of them you have implicitly allowed order to be important. That is, you are essentially saying that choosing items 4 and 5 is different than choosing items 5 and 4.

We would divide by 2! in this case since order is not important. That is for identical objects 45 is the same as 54. The 2!, of course, is the number of ways that 2 items can be ordered. All of those ways actually count for just one (when order is unimportant).

Same thing is happening here when heehaww said the numerator was 3*4*5*9. (Multiplying that last "*9" is the analog to the multiplying by 7 in the 8*7 above.)

P.S. Perhaps it is more obvious if you split the calculation into the three cases (A repeats or B repeats or C repeats).

For then it is clear than the numerator is ((3*2)/2)*4*5 + 3*((4*3)/2)*5 + 3*4*((5*4)/2). Clearly in each term you need to divide by 2 for the reason stated above (the order of choosing identical objects cannot matter). This simplifies to (3*4*5)*(2+3+4)/2 or (3*4*5*9)/2. Of course, heehaww's logic performs all of these calculations in one step as your efficiency mandate required.

Someone could take
• two As, one B and one C,
• one A, two Bs and one C, or
• one A, one B and two Cs.
There are just those three ways to do it.

There are
• 3 ways to take one A and 3 ways to take two As,
• 4 ways to take one B and 6 ways to take two Bs,
• 5 ways to take one C and 10 ways to take two Cs,

Thus there are
3*4*5+3*6*5+3*4*10=270 possible ways to have at least one A, at least one B, and at least one C.

But all in all, the number of ways to choose 4 of 12 objects is: C(12,4)=495.

Thus the probability is 270/495=0.545455

I don't know what the best way to solve the problem. I guess look and see who got that answer in the shortest number of steps. (I haven't looked at the other answers yet).

[Edit] OK. Now I looked. HeeHaww and masque de Z both got that answer faster than I did. (6/11=0.545455). [/Edit]

Buzz

Whosnext took the words out of my mouth. Does that help you, Sparky?

Ofc if we use permutations we don't have to divide by the 2!. Here it is with perms:

3*4*5*9*(4*3) / (12*11*10*9) = 60/(11*10) = 6/11

The denominator counts all orders, and all orders are valid so the numerator must count all orders too. We've established that 60*9 counts the 2 orders of the repeated symbol, so what's missing are the ways to arrange, say, ABCC. There are 4 possible spots for the A and 3 for the B, so 12. Or C(4,2) for the C's and then 2! ways to arrange the A&B so again 12. Or you don't have to count it at all, you just know that the total orders should be 4! and that you've already counted 2! (via the 60*9), so multiply by 4!/2!.

Thank you for all the great explanations! Really helpful!