## Introduction to Permutations

Problem: In how many ways can Anna, Brenda and Connie stand in a single line for picture taking?

Intuitively, we can count the number of ways by listing. We can list randomly as shown below.

Anna, Brenda, Connie

Brenda, Connie, Anna

Connie, Anna, Brenda

and so on.

Q1: Do you think that listing randomly is a good idea? What are its advantages and its disadvantages?

Listing randomly can solve our problem, if there are only a few things, or in our case persons, to be arranged; however, we can do better than that.  Learning mathematics has taught us to be organized, and has taught us to do things systematically.  Besides, if there are many persons to be arranged, it is hard to keep track if we have listed all possible arrangements. For example, what if David joins the group? Try to list randomly and determine how many possible arrangements are there.

Q2: Before proceeding, can you think of a way to come up with an organized way to list all the possible arrangements?

One possible strategy is to list in alphabetical order. Let us represent Anna, Brenda and Connie by the first letter of their names. If we choose A to occupy the leftmost position, then there are two possible choices for the middle position, namely B and C. That means have AB and AC as all possible arrangements if A is chosen to occupy the leftmost position. Now, in each of the cases, we only have one person left to occupy the rightmost position.  This gives as ABC and ACB as all possible arrangements of the three girls if A were to occupy the leftmost position.

We can also use a tree diagram as shown in Figure 1.  If we choose A to be the person in leftmost position, then the branches B and C mean our possible choices for the middle position. If we have chosen a person who will occupy the middle position, then we are left with only one person to occupy the rightmost position.  Hence, if we choose A to occupy the first position, the only possible arrangements for picture taking are ABC and ACB.

Figure 1- The tree diagram of all the possible arrangements of A, B and C.

But we know that we can also choose B or C as the person who will occupy the leftmost position. This means, that there are 3 possible choices for the first position, 2 possible choices  for the second position and 1 possible choice for the third position (see Figure 1). Hence, there are 3 x 2 x 1 possible arrangements for 3 persons in a single line.

We will denote 3 x 2 x 1 as 3! (3 factorial).  This implies that if we say 5!, we mean 5 x 4 x 3 x 2 x 1 = 120.  In general, n! means n(n-1)(n-2)…(3)(2)(1). Note that the ellipsis symbol … denotes that there are numbers in the sequence that are not shown.  For example, 100(99)(98)…(3)(2)(1), means the product of all integers from 100 all the way down to 1.

Q3: If David joins the group, how many possible arrangements are there?

You may want to solve this problem first before proceeding.

Figure 2 – The tree diagram of all possible arrangments of A, B, C and D.

Looking at the tree diagram, there are four possible choices to occupy the leftmost position, 3 possible choices to occupy the second position, 2 possible choices to occupy  the third position and 1 possible choice to occupy the rightmost position.  Hence there are 4! = 4 x 3 x 2 x 1 = 24 possible arrangements.

By now, you would have realized that the number of arrangements or the number of permutations of n persons on a single line for picture taking is n!. We will denote it as P(n,n) orthe permutations of n objects taken n at a time. We sayn objects taken n at a time because we have the choice to choose numbers less than n to be arranged.  For example, we can choose A and C from A, B, C and D .  This means that we a permutation of 4 objects taken 2 at a time. In general, we describe this type of permutation as permutations of n objects taken k at a time and write P(n,k).

Let us see what happens to our computation with P(4,2). Since there are 4 possible choices for the first choice, and 3 choices for the second position, therefore, there are 4 x 3 possible permutations.  This is the same as removing the smallest two factors by division. If we do this, we come up with the following computation:

If we list the elements of P(4,2), we have the following: AB, BA, AC, CA, AD, DA, BC, CB, BD, DB, CD, and DC. Indeed, we have 12 possible arrangements.

With our findings above, let us try to perform a few more computations and see if we can find a pattern.

Therefore, by looking at the pattern, we can conclude that the number of permutations of n things taken k at a time described by the formula

You may want to read  Introduction to Combinations, the continuation of this post.

## Why is the angle sum of the interior angles of a triangle 180 degrees?

We were taught that the sum of the measures of interior angles of a triangle is 180 degrees. But how come? Is it true no matter what the shape or size of the triangle?

Figure 1 – Triangles with interior angles a, b and c

Recall that an angle is the amount of rotation of a ray.  In Figure 2, the ray was rotated from A to C, and the amount of rotation is 60 degrees. We can say that the measure of angle ABC is 60 degrees.

Figure 2 – Ray containing A is rotated to to C making a 60-degree angle rotation

The rotation from A to D forms a straight line and measures 180 degrees.  Therefore, straight angle ABD measures 180 degrees.  It follows that a 180-degree rotation is a half-circle. Therefore, a complete rotation is 360 degrees.

We can verify if our question about the sum of the interior angles of a triangle by drawing a triangle on a paper, cutting the corners, meeting the corners (vertices) at one point  such that the sides coincide with no gaps and overlaps (see Figure 3). Notice that no matter what the size or shape of a triangle, as long as the previous conditions are met, the two of its sides will be collinear as shown in the Figure below.

Figure 4 – The corners of a triangle put together

However, this is not the proof. To discuss the proof, we are going to use Euclid’s fifth postulate. Euclid’s fifth postulate tells us that if a parallel line is cut by a transversal, their corresponding angles are congruent. In the diagram below, lines p and q are parallel lines and angles shown with the same letters are corresponding angles, and hence congruent.

Figure 5 – Parallel lines p and q cut by a transversal.

Now how do we connect Parallel Postulate to the angle sum of a triangle. Consider a triangle with interior angles a, b, and c. Using the fifth postulate, we use the side containing angles a and c as base of the triangle and extend it to both sides. Next, we draw a line to the base and passing thorugh the vertex not on the base.

Figure 6 – Triangle with three points on two parallel lines

Now, the sum of angles a, b and c is 180 degrees (why?). But a, b and c are also the interior angles of a polygon. Therefore, the sum of the interior angles of a triangle is 180 degrees.

You may also want to read Sum of the Interior Angles of a Polygon.

## Why is any number raised to 0 equals 1?

Why is any number raised to 0 equals 1?

If we raise a number to an exponent, we are multiplying it by itself a certain number of times.  For example, $3^4$ means you have to multiply $3$ by itself $4$ times.   In exponential notation, we call $3$ the base and $4$ the exponent.

Shown below are examples of exponential expressions and their expansion.

$2^3 = 2 \cdot 2 \cdot 2$

$4^7 = 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4\cdot 4\cdot4$

$x^4 = x \cdot x \cdot x \cdot x$

$(a + b)^3 = (a+ b)(a+b)(a+b)$

$x^m = x \cdot x \cdot x \cdot \ldots \cdot x \cdot x$ (Multiply $x$ by itself, $m$ times)

The$\ldots$ symbol means “and so on.” It represents $x$’s that are missing.  It is convenient to use the said symbol for large values of $m$.

Multiplying Expressions with Exponents

If we want to multiply expressions with the same base, let us see what happens. For example, what will happen if we multiply $2^3$ and $2^5$?

From above, $2^3 = 2 \cdot 2 \cdot 2$ and $2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$. Multiplying the two expressions, we have

$2^3 \cdot 2^5 = (2 \cdot 2 \cdot 2)(2 \cdot 2 \cdot 2 \cdot 2 \cdot 2) = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^8$

From our computation, we can conclude that if we multiply to expressions with the same base, we have just have to add their exponents (Can you see why?). That is, for expressions $x^m$ and $x^n$,

$x^m \cdot x^n = x^{m+n}$ (*)

Q1: What if the base of the two expressions are not the same? Will our formula above still apply?

Dividing Expressions with Exponents

What about dividing expressions with exponents? Suppose, we want to divide $2^5$  by $2^3$.

We know that $2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ and $2^3 = 2 \cdot 2 \cdot 2$. Dividing the two expressions, we have

$\displaystyle \frac{2^5}{2^3}= \displaystyle\frac{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2}{2 \cdot 2 \cdot 2} = 2^2$

Since, three $2$‘s are canceled out, we can therefore conclude that in dividing two expressions with the same base, we just have to subtract their exponents. That is, for expressions $x^m$ and $x^n$,

$\displaystyle\frac{x^m}{x^n} = x^{m-n}$. (**)

What happens if the exponent of the denominator is larger? For example, $\displaystyle \frac{2^2}{2^7}$?

From (**), $\displaystyle\frac{2^2}{2^7} = 2^{2-7} = 2^{-5}$.

Now, let us compare this result when we expand our expression:

$\displaystyle \frac{2^2}{2^7}= \displaystyle\frac{2 \cdot 2}{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2} = \frac{1}{2^5}$

Our observation tells us that, $2^{-5} = \displaystyle\frac{1}{2^5}$. Therefore, $x^{-m}= \displaystyle\frac{1}{x^m}$.

Q2: In general, what is the value of $\frac{x^m}{x^n}$ if $m < n$?

What if the exponents of the numerator are equal? For instance, $\frac{2^5}{2^5}$. This is practically dividing the same number, so obviously the answer is $1$. However, we can also use our conclusion above.

From (**), $\displaystyle\frac{2^5}{2^5} = 2^{5-5} = 2^0 = 1$

Conclusion

Here we observe that raising$x$ (or any expression) to $0$  means that the number of factors in the numerator and the number of factors in the denominator is the same. Therefore, $x^0$ can be expressed as $x^{m-m}$ for any value of $m$. But from (**), $x^{m-m}$ is equivalent to $\displaystyle\frac{x^m}{x^m} = 1$

Therefore, any number raised to $0$ (with the exception of 0) equals $1$.

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