Here is an example to something similar I saw in one of the
OG or
Kaplan books that I can show the difference between a combination and a permutation. Two versions of the question.
Permutation Version:There are 11 chemicals jars in a laboratory that are identified by a two color labeling system. How many colors are needed to identify all 11 chemicals? The answer is: 4 colorsQuick solution....
-4 labels using the same color twice
-11-4= 7 labels remaining that will use two different colors
-4*3 = 12 additional color combinations so this is sufficient
Longer solution...
-You know you will need significantly less then 11 colors to do this so pick a number like 3 to start with.
-three colors(R, G, and B) will give you three labels with: RR, GG, and BB
-You still need 11-3 = 8 labels.
-The resulting possibilities of mixing the three colors are: RG, RB, GR, GB, BR, and BG = 6 labels
-3+8 = 11. Not enough possibilites.
-So, 3 is barely too small. Work up from there. Let's try 4.
-four colors(R, G, B, and Y) will give you four labels with: RR, GG, BB, and YY
-You still need 11-4 = 7 labels.
-The resulting possibilities of mixing the four colors are: RG, RB, RY, GR, GB, GY, BR, BG, BY, YR, YG, and YB = 12 labels
-4+12 = 16 label possibilities which is sufficient.
Combination Version:There are 11 chemicals jars in a laboratory that are identified by a two color labeling system. How many colors are needed to identify all 11 chemicals if the order on the label doesn't matter? The answer is: 5 colorsQuick solution....
-5 labels using the same color twice
-11-5= 6 labels remaining that will use two different colors
-(5*4)/2 = 10 additional color combinations.
-10+5 = 15 total possibilities with 5 colors.
Longer solution...
-The same approach applies so we will start out with 4 colors.
-four colors(R, G, B, and Y) will give you four labels with: RR, GG, BB, and YY
-You still need 11-4 = 7 labels remaining
-The resulting possibilities of mixing the four colors are: RG, RB, RY, GR, GB, GY, BR, BG, BY, YR, YG, and YB = 12 labels
-Since the problem said that color order on the label doesn't matter so RG is the same as GR. We have to eliminate the duplicates. 12/2 = 6 possibilities.
-4+6 = 10, which is not sufficient.
-Let's try five colors (R, G, B, Y, and O). These will give you: RR, GG, BB, YY, and OO = 5 labels.
-11-5 =6 labels remaining.
-The resulting possibilities of mixing the four colors are: RG, RB, RY, RO, GR, GB, GY, GO, BR, BG, BY, BO, YR, YG, YB, YO, OR,
OG, OB, and OY = 20 possibilities.
-Again, since the problem said that color order on the label doesn't matter we have to eliminate the duplicates. 20/2 = 10.
-10+5 = 15 total label possibilites from 5 colors which is sufficient.
Another example:
Permutation Version:There are 4 teams in a league. How many different games are possible?4*3 = 12 games possible.
12, 13, 14, 21, 23, 24, 31, 32, 34, 41, 42, 43
Combination Version: There are 4 teams in a league. How many different games are possible if two teams only play each other once?(4*2)/2 = 6 games possible.
12, 13, 14, 23, 24, 34
For combinations you are just getting rid of duplicates.
Hope this helps.