Discrete Math (Permutation or Combination)?

Question

Textbook Solution:

a) $2^5$

b) $2^n$ is the general formula.

Logical Questions:

1. For this quesetion, how do you know if order matters and repetitions are allowed ?

Answer 1

I though about your answers and this problem, and I decided to simplify what I think the solution would be.

For every n distinct object, there are two choices. Either itself or not itself. When it is not itself, then you can choose an identical coin.

Imagine you have 6 coins. And 3 buckets to put 1 coin per each.

1 2 3 A A A --> your coins

I___I I___I I___I --> your buckets

1, 2, and 3 are three distinct coins. A, A, and A are clearly identical coins.

You end up with these selections:

1 2 3

1 2 A

1 3 A

1 A A

A A A

2 3 A

2 A A

3 A A

= 8 selections

Formula $2^3$ = 8

That means for every bucket, you can make "two choices. Either itself or not itself. When it is not itself, then you can choose an identical coin".

$\therefore$ if given 5 coins, and 5 buckets. You have $2^5$ choices.

Answer 2

So in the situation that we are faced with, we have $n$ distinguishable objects, and $n$ undistinguishable objects. We have to pick $n$ of these objects.

We split into cases:

case $1$: We pick all the distinguishable objects. This is done in $1$ way.

case $2$: We pick $n-1$ distinguishable objects and $1$ undistinguishable object. This is done in $n$ ways (picking an object out of the $n$ distinguishable ones is done in $n$ ways).

case $3$: We pick $n-2$ distinguishable objects and $2$ undistinguishable objects. This is done in $\binom n{n-2}$ ways (we pick $n-2$ distinguishable objects out of $n$ , and because the other category of objects is indistinguishable, and we have to pick exactly $2$ of these objects, there's no choice: only one possibility).

I would like to point out above that because we are selecting objects and not arranging them, we are taking combinations and not permutations.

case $k$: We pick $n-k$ distinguishable objects and $n$ undistingishable objects. For the reason mentioned above, we have that this can be done in $\binom n{n-k}$ ways.

Now, $k$ goes from $0$ to $n$, and all the given cases are disjoint. So, we have to calculate the sum $\displaystyle\sum_{k=0}^{n} \binom n{n-k}$. This, and you should remember your binomial theorem here, is just $2^n$. Hence, the answer is $2^n$, and for $n=5$ (the dime, nickel, penny ,quarter,half-dollar are distinguishable objects, whereas the Susan Anthony dollars aren't) the answer is $2^5$.

Do get back on doubts!

Answer 3

1. For this quesetion, how do you know if order matters and repetitions are allowed ?

There is no simple answer to this that make you understand WHY there are repetitions or not, or WHY order matters or not. The best thing you can do is make the experiment yourself (took some coins and see how many different groups you can do, start with 3 coins, after 4, etc...), then you can understand why in some situation something have repetitions or order matters.

Otherwise you will not learn something.

Answer 4

In general, if you're simply asked for the number of ways to select $k$ objects, then order doesn't matter. All that matters is whether each object is selected or unselected. It doesn't make a difference whether you select a dime then a penny or a penny then a dime, because you will still end up with the same two coins.

If order matters, then the question needs to make that clear. That can be explicit ("In how many ways can we choose three people, in order, from ten?") or implicit ("I want to award three prizes of different values to three different people from a set of ten; in how many ways can I do this?"). In the second example there it is the fact that the prizes are different which tells you that order matters -- if you give the prizes to the same people in a different order, then they will get different amounts.

The question you ask has an extra complication, that some of the objects are identical, i.e. choosing the dime and the first Susan B. Anthony dollar is the same as choosing the second Susan B. Anthony dollar and the dime. Either way you'll end up with the same two coins, so far as you can tell.

For a simple way to see why the formula in your textbook works, note that you can choose whether to select the penny or not, then choose whether to select the nickel or not, then the dime, then the quarter, then the half-dollar. Once you've done that, you have from $0$ to $5$ coins, and there is exactly one way to make it up to exactly $5$ coins by adding some (possibly $0$) Susan B. Anthonys. Therefore there are $2^5$ choices, since for each of the five smaller coins you have $2$ options. Exactly the same argument works for $n$ different and $n$ identical objects.