Combinations
There are also two types of combinations (remember the order does not matter now):
- Repetition is Allowed: such as coins in your pocket (5,5,5,10,10)
- No Repetition: such as lottery numbers (2,14,15,27,30,33)
1. Combinations with Repetition
Actually, these are the hardest to explain, so I will come back to this later.
2. Combinations without Repetition
This is how lotteries work. The numbers are drawn one at a time, and if you have the lucky numbers (no matter what order) you win!
The easiest way to explain it is to:
- assume that the order does matter (ie permutations),
- then alter it so the order does not matter.
Going back to our pool ball example, let us say that you just want to know which 3 pool balls were chosen, not the order.
We already know that 3 out of 16 gave us 3,360 permutations.
But many of those will be the same to us now, because we don't care what order!
For example, let us say balls 1, 2 and 3 were chosen. These are the possibilites:
| Order does matter | Order doesn't matter |
| 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1 | 1 2 3 |
So, the permutations will have 6 times as many possibilites.
In fact there is an easy way to work out how many ways "1 2 3" could be placed in order, and we have already talked about it. The answer is:
3! = 3 × 2 × 1 = 6
(Another example: 4 things can be placed in 4! = 4 × 3 × 2 × 1 = 24 different ways, try it for yourself!)
So, all we need to do is adjust our permutations formula to reduce it by how many ways the objects could be in order (because we aren't interested in the order any more):
That formula is so important it is often just written in big parentheses like this:
 |
| where n is the number of things to choose from, and you choose r of them (No repetition, order doesn't matter) |
It is often called "n choose r" (such as "16 choose 3")
And is also known as the "Binomial Coefficient"
Notation
As well as the "big parentheses", people also use these notations:
Example
So, our pool ball example (now without order) is:
| 16! | = | 16! | = | 20,922,789,888,000 | = 560 |
| 3!(16-3)! | 3!×13! | 6×6,227,020,800 |
Or you could do it this way:
| 16×15×14 | = | 3360 | = 560 |
| 3×2×1 | 6 |
So remember, do the permutation, then reduce by a further "r!"
... or better still ...
Remember the Formula!
It is interesting to also note how this formula is nice and symmetrical:
In other words choosing 3 balls out of 16, or choosing 13 balls out of 16 have the same number of combinations.
| 16! | = | 16! | = | 16! | = 560 |
| 3!(16-3)! | 13!(16-13)! | 3!×13! |
1. Combinations with Repetition
OK, now we can tackle this one ...
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Let us say there are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be?
Let's use letters for the flavors: {b, c, l, s, v}. Example selections would be
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(And just to be clear: There are n=5 things to choose from, and you choose r=3 of them.
Order does not matter, and you can repeat!)
Order does not matter, and you can repeat!)
Now, I can't describe directly to you how to calculate this, but I can show you a special technique that lets you work it out.
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Think about the ice cream being in boxes, you could say "move past the first box, then take 3 scoops, then move along 3 more boxes to the end" and you will have 3 scoops of chocolate!
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| So, it is like you are ordering a robot to get your ice cream, but it doesn't change anything, you still get what you want. |
Now you could write this down as 
(arrow means move, circle means scoop).
(arrow means move, circle means scoop).
In fact the three examples above would be written like this:
| {c, c, c} (3 scoops of chocolate): | |
| {b, l, v} (one each of banana, lemon and vanilla): |  |
| {b, v, v} (one of banana, two of vanilla): |  |
OK, so instead of worrying about different flavors, we have a simpler problem to solve: "how many different ways can you arrange arrows and circles"
Notice that there are always 3 circles (3 scoops of ice cream) and 4 arrows (you need to move 4 times to go from the 1st to 5th container).
So (being general here) there are r + (n-1) positions, and we want to choose r of them to have circles.
This is like saying "we have r + (n-1) pool balls and want to choose r of them". In other words it is now like the pool balls problem, but with slightly changed numbers. And you would write it like this:
 |
| where n is the number of things to choose from, and you choose r of them (Repetition allowed, order doesn't matter) |
Interestingly, we could have looked at the arrows instead of the circles, and we would have then been saying "we have r + (n-1) positions and want to choose (n-1) of them to have arrows", and the answer would be the same ...
So, what about our example, what is the answer?
| (5+3-1)! | = | 7! | = | 5040 | = 35 |
| 3!(5-1)! | 3!×4! | 6×24 |