# A bag contains four identical white balls, three identical red balls, and three identical blue balls.

A bag contains four identical white balls, three identical red balls, and three identical blue balls.

A) How many ways are there to choose three balls in order?

B) How many ways are there to choose three balls ignoring order?

If i was to guess on how to do these problems i would think C(10,3) is part of the equation but i really don't understand, if someone could tell me simply how to look at these problems it would be very helpful.

• Do you put a ball back in the bag before choosing the next one? – graydad Mar 24 '15 at 16:24
• it's involving repetition so i'm assuming that means to put the ball back? – Charlene Mar 24 '15 at 16:25
• The answer to A) is 27 if that helps, i honestly don't know – Charlene Mar 24 '15 at 16:27
• no, i dont even understand what "in order" or "ignoring order" even means in this context – Charlene Mar 24 '15 at 16:31
• Order matters would mean that drawing WRB is different than drawing BRW. If order didn't matter, those two draws would be considered the same. – graydad Mar 24 '15 at 16:34

## 2 Answers

A) Deals with permutations:

$\frac{n!}{(n-r)!}$

B) Deals with combinations, like you suggested:

$\frac{n!}{r!(n-r)!}$

Where n is the number of balls in the bag, and r is the size of the group, namely 3.

The key difference is that order matters in permutations. Of course, if you start getting into "putting balls back" and such, it becomes a little more intricate.

• If order matters in permutations, wouldn't $B$ not deal with permutations since it says "ignoring order"? – graydad Mar 24 '15 at 16:36
• welcome @user162393 get a good screen name :D – RE60K Mar 24 '15 at 16:36
• Thank you graydad, I mixed up the cases. It's been corrected. – AlexMayle Mar 24 '15 at 17:06

For A suppose you take 3 balls out and arrange them in a stand, like the one for first three rankings after an olympics event, first place can go to any of the three colours, similiarly the rest two to 3 and 3 because the count of every ball is greater than 3. Using product rule $3.3.3=27$ For B take 3 balls out and put in a bag, which is first and which is last no one knows, there are 3 balls in total, let the no of balls of each colour be deoted by $x_i$, and the sum of these is 3 so $x_r+x_w+x_b$ where $0\le x_i\le 3<4$ because even if there are 4 white balls you can't take all of them out and now we use stars and bars because even though upper limit is 3 it is greater than or equal to RHS, i.e. 3 (here it is equal), so it is $^{3+3-1}{\rm C}_{3-1}={}^{5}{\rm C}_2=10$