Ordered selection (Permutation)

Suppose there are 10 songs in your music collection and you want to shuffle play them all (i.e. the 10 songs would play in a random order). In how many different ways can the collection be played?

This type of problem is called ordered selection (permutation), as we are selecting the songs from the collection in a certain order. For instace, these selections are considered different:

${\displaystyle {\begin{array}{l}1,2,3,4,5,6,7,8,9,10\\7,3,10,4,9,5,1,8,2,6\\2,1,3,4,5,6,7,8,9,10\end{array}}}$

These clearly define different orders or in effect, playlists.

Let's think this step by step. We can pick 10 different songs for the first position (or first song to be played), 9 for the second position (as there is 1 song already picked), 8 for the third (as 2 songs are fixed), and so on and so forth. The total number of ways can be calculated as follows:

${\displaystyle 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}$

Almost 4 million different playlists!

Now we shall introduce the factorial function, which is a compact way of expressing:

${\displaystyle n!=n\times (n-1)\times ...\times 3\times 2\times 1}$

Formally defined by:

${\displaystyle {\begin{array}{l}0!=1\\n!=n\times (n-1)!\end{array}}}$

So in our case:

${\displaystyle {\begin{array}{l}10!=10\times 9!=10\times 9\times 8!...\\10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1\end{array}}}$

What if we had 30 songs, but still wanted to shuffle just 10 of them? Similar reasoning leads to

${\displaystyle 30\times 29\times 28\times 27\times 26\times 25\times 24\times 23\times 22\times 21}$

different ways. This is equivalent to:

${\displaystyle {\frac {30!}{20!}}={\frac {30!}{(30-10)!}}}$

Generally speaking, the number of permutations of m items out of n (for instance, 10 songs out of 30) is:

${\displaystyle {\frac {n!}{(n-m)!}}}$

The idea behind it is that we cancel out all but the first m factors of the n! product:

${\displaystyle 30!=\underbrace {30\times 29\times ...\times 22\times 21} _{\text{First 10 factors}}\times \underbrace {20\times ...\times 2\times 1} _{(30-10)!=20!}}$

When m equals n we say we are counting the number of permutations of n items:

${\displaystyle {\frac {n!}{(n-n)!}}={\frac {n!}{0!}}=n!}$

While solving counting problems, we'll generally use the words items or elements (in this case, our songs) and sets (in this case, the collection or playlists).

1. In how many ways can I arrange my bookshelf (7 books) if I own a trilogy and the 3 books have to be always together, in the same order?

2. In how many ways can I arrange my family (6 members) in a line, with the only rule that my parents have to be in adjacent positions?

1. How many three-digit numbers can be formed with only prime digits, and odd digits may only be repeated twice?

1. What if there are 7 people, but the table only has 5 chairs?

Unordered selection (Combination)

Out of the 15 people in the math class, 5 will be chosen to represent the class in a school wide mathematics competition. How many ways are there to choose the 5 students?

This type of problem is called an unordered selection (combination), as the order in which you select the students is not important. For example, these selections are considered equivalent:

${\displaystyle {\begin{array}{l}{\text{Joe, Lee, Sue, Violet, Justin}}\\{\text{Lee, Joe, Sue, Justin, Violet}}\end{array}}}$

As seen before, there are

${\displaystyle {\frac {15!}{10!}}}$

ways to choose the 5 candidates in ordered selection, but there are 5! ordered selections for each different combination (i.e. the same 5 students). This means we are counting each combination 5! times. Consequently, there are

${\displaystyle {\frac {15!}{10!\times 5!}}}$

ways of choosing 5 students to represent the class.

In general, the number of ways to choose m items out of n items is:

${\displaystyle {\frac {n!}{(n-m)!\times m!}}}$

We took the formula for permutations of m items out of n, and then divided by m! because each combination was counted as m! ordered selections (m! times).

This formula is denoted by:

${\displaystyle {n \choose m}={\frac {n!}{(n-m)!\times m!}}}$

Note that ${\displaystyle {n \choose m}}$ is often read as n choose m.

1. Try not to use a calculator and think of the problem with groups of 1 student.

2. Again, no calculator and this time there are 1 million students in the class and we want to form groups of 999,999 students.

Binomial coefficient

The binomial coefficient

${\displaystyle {n \choose m}}$

expresses the number of unordered groups that can be formed by choosing m elements from a set containing n.

These are some basic and useful properties:

• The number of ways of choosing m elements out of n equals the number of ways of (not) choosing the remaining n - m out of the n elements. A group of selected elements defines another group of non selected elements.

${\displaystyle {n \choose m}={n \choose {n-m}}}$

• There is only one way to choose no elements, or to (not) choose them all.

${\displaystyle {n \choose 0}={n \choose n}=1}$

• There are n ways of choosing one element, or (not) choosing n - 1.

${\displaystyle {n \choose 1}={n \choose {n-1}}=n}$

• The difference between

${\displaystyle {n \choose m}}$

and

${\displaystyle {{n+1} \choose m}}$

is that in the first formula we are not counting the groups which include the element added in the second formula. The amount of groups size m which include a particular element from a set with n + 1 elements is

${\displaystyle {{(n+1)-1} \choose {m-1}}={n \choose {m-1}}}$

therefore

${\displaystyle {{n+1} \choose m}={n \choose m}+{n \choose {m-1}}}$

• The difference between

${\displaystyle {n \choose {m+1}}}$

and

${\displaystyle {n \choose m}}$

is that in the first formula we are counting groups with one more element than the ones in the second formula. Every group size m needs one "missing" element of the n - m remaining in order to grow to size m - 1.

${\displaystyle {n \choose m}\times (n-m)}$

However, we must observe that we are counting every group size m + 1 many times, m + 1 times indeed. This is because the "missing" element can be any of the m + 1 elements in the resulting group, in other words, m + 1 groups size m (when associated with the right element) will formed the same group size m + 1. How many groups size m can we formed by removing an element from a group size m + 1?

${\displaystyle {{m+1} \choose 1}=m+1}$

This suggests

${\displaystyle {n \choose {m+1}}={n \choose m}\times {\frac {n-m}{m+1}}}$

Exercises