Some combinatorics questions...?

Question

You have 4 letters (A,B,C,D) and 2 boxes, one red and one blue.  

a) You want to divide the letters, 2 per box.  How many ways can you do this?
b) How would your answer to part (a) differ if both boxes were red?

Now consider six letters (A-F), and 3 boxes (red, blue, yellow).

a) How many ways can we divide up the letters among the three boxes, 2 letters per box?
b) How would your answer to part (a) change if all 3 boxes were one color?

The Zeitgeist · Accepted Answer

(Let's see how much combi I remember)

1a) Start by placing the A in the red box.  Then you have three choices for the other ball in the red box.  These three choices completely describe the possibilities with A in Red.  But you could have put A in blue, for another three possibilities.  So there are six possible ways.

1b)  If both boxes were red, you would only have the three ways to arrange the letters because, say AC, BD is now the same as BD, AC.


2a)  With 6 letters in 3 boxes, we again start by placing A in the red box.  Then there are 5 possible letters to go in the red box with A.  For example, let's choose D.  Then there are 6 ways to put B,C,E,F into the blue and yellow boxes (see 1a).  So for each of the 5 letters you can put with A, there are 6 ways to go, thus there are 5*6 = 30 ways once A is placed.  But there are 3! = 6 different ways to arrange one grouping of letters in the colored boxes.  For example, if you had gotten Red: AD, Blue: CF and Yellow: BE, then we could switch the boxes around to get 5 more possible arrangements.
RBY
RYB
BRY
BYR
YBR
YRB
(keeping the letters the same (AD)(CF)(BE) ).

Thus, there are 120 ways.

2b)  If all three boxes were the same color, you wouldn't multiply by 3! , so you would have 20 possible ways.

Hope this helps.

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Some combinatorics questions...?

1 Answer