# Counting Basics

**Fundamental Principle of Multiplication:**

Suppose that you are making a sandwich. If there are **3** types of bread to choose from, and there are **7** different condiments that you can put in the sandwich, the total number of sandwiches you can make is **3*7 = 21**.

This can be generalized to any number of choices. If a task can be done in **m** ways, and the next task can be done in **n** ways, then the two jobs together can be done in **m*n** ways.

**Example 1:**

**Q: Amy has three skirts and five blouses. How many skirt-blouse outfits can Amy wear**?

**Solution**: Amy has three choices for skirts and five choices for blouses. For each of the choice of her skirt, Amy has five choices for a blouse to make an outfit. So she can make

**3 X 5 = 15**

Skirt-blouse outfits.

** ****Concept of Permutation:**

** **The word permutation means arrangement. Think about this as how many ways we can arrange something. So if we have three people **A, B** and **C** trying to sit in a movie theater, how many ways can they be seated?

Permutations: **ABC, ACB, BAC, BCA, CAB, CBA**

There are **6** ways to do this. Now let’s see how we can do this without having to write them all out.

**Example 2:**

**Q: In how many ways can 10 people be seated in a row of 10 chairs?**

*Solution**:*

We can use this example to calculate permutations in a fast and easy way. We can choose any of the **10** people to sit in the first seat, so we have **10** choices. Now there are **9** people left to choose from, so for the second seat, there are **9** choices. Similarly, there are **8** choices for the third seat, **7** for the fourth seat and so on until there is only one choice for the last seat. Thus, the total number of choices, using our rule of multiplication from above is **10*9*8*7….*3*2*1**.

This number can also be written as **10!**, and it is called a factorial. In general, **n!** is the same as **n*(n-1)*(n-2)*…*3*2*1**. Essentially, it is the same as multiplying all the numbers equal to or less than **n**. A special case for factorial occurs when you consider **0!**, and this value happens to be **1** (since there is only one way to count nothing!)

Let’s look at another example:

**Example 3:**

**Q: In how many ways can a row of 5 seats be filled from a set of 12 people?**

*Solution**:* Similar to the previous problem, we have **12** choices for the first seat, **11** for the second, **10** for the third, **9** for the fourth and **8** for the fifth seat. So the total number of ways is

** 12 x 11 x 10 x 9 x 8**

It is important to note that the order of the people sitting is important in a permutation. Also, we can’t have any repetitions. This particular permutations is one of **12** objects(number of people) taken **5** at a time (seats to be filled).

Now, let’s look at a case where we might be over-counting our choices.

**Over counting:**

**Example 4:**

**Q: How many different ways can the five letters of the word STATE be scrambled?**

**Solution**: As we have seen earlier, five letters of the word **STATE** can be scrambled in **5*4*3*2*1 = 5!** ways.

However, note that the letter **“T”** is repeated twice and these two **T**’s are indistinguishable and provide the same arrangement during scrambling. So we over counted the number of ways of scrambling this word. These two **T**‘s can be arranged in **2!** ways. To make the counting correct, we want to divide our total by **2!**.

So the total arrangements is **5!/2! = 120/2 = 60** ways.

Practice makes perfect, so let’s try another example:

**Example 5:**

**Q: How many different ways can the five letters of the word BREEZE be scrambled?**

**Solution**: There are 6 letters, so total number of permutations is **6*5*4*3*2*1 = 6!**. However, we see that **“E”** is repeated three times, and they are over counted by **3*2*1 = 3!** times. So, the different ways of scrambling **BREEZE** is

**6!/3! = 120** ways

So we’ve seen permutations, and permutations with repetitions thus far. Now, let’s move on to cases where the order doesn’t matter. These are called combinations. For example, when we pick **5** people for a team, we don’t care what order they are picked, we only care about which five people we have on the team. Let’s look at the mathematics behind this idea of combinations:

**Combination:**

**Example 6:**

**Q: In how many ways can a team of 5 people be selected from a group of 12 people?**

Note that the **order is not important** here. It is immaterial the way **5** people are selected. Arrangements in which order is not important are called combinations. We say that this is a combination of **12** objects (number of people) taken **5** at a time (seats to be filled).

*Solution**:* If we first think of how many ways we can pick the **5** people, there are **12** choices for the first person, **11** for the second, and so on. Thus like before, we get a total of **12*11*10*9*8**. However, since order does not matter, any of the five permutations of selected people would produce the same combination. What do we mean by this? Suppose we select Amy, Bush, Clara, Danny and Eve. The order in which they are selected doesn’t matter. There are **5*4*3*2*1 = 5!** ways to arrange them. All these arrangements still result in the same **5** people being chosen. Thus, we are over-counting. Therefore, we need to divide the total arrangements by **5!**.The final answer becomes **(12*11*10*9*8)/5!**.

Let’s look at another example of combinations:

**Example 7:** * ***A merchant decides to put two of his eight VCR models on sale, but has not decided which two. How many different selections can she make?**

**Solution:** The first one can be picked in **8** ways, and the second in **7** ways. However, we don’t care about the order, so we divide by **2!**. Thus the total is **8*7/2! = 28** ways.

**Quick Recap:**

So far, started with the simple idea of multiplication, and counted permutations or arrangements of objects. Then we looked at permutations with repetitions. Finally, we looked at combinations, where the order doesn’t matter when counting objects. Now let’s look at one last way to count the something. By counting what we **don’t** want, and subtracting that amount from the total number of options. This is called **complimentary counting**.

**Complementary Counting**

**Example 8:**

**Q: How many three-digit numbers are not multiple of 7?**

Since the title above says complimentary counting, I guess we can give it a shot! Let’s count how many three digit numbers **are** a multiple of **7**, and subtract that from the total number of three digit numbers.

The first multiple of **7** that is a three digit number is **7*15 = 105**. The last one is **7*142 = 994**. This means all the multiples of **7** from **15** to **142** are the ones we are looking for. There are **128** of these. Now, there are **900** total three digit numbers (since **100** is the lowest, and **999** is the highest). Thus, the total number of three digit number that are not a multiple of **7** is **900-128 = 772**.

Let’s look at a slightly more complicated problem which can be solved by complimentary counting:

**Example 9:**

**Q: The Smith family has 4 sons and 3 daughters. In how many ways can they be seated in a row of 7 chairs such that at least 2 boys are next to each other?**

Solution: So, we want to first count the number of ways that the **7** children can be seated such that no two boys are seated together. We can then subtract this from the total number of ways that the **7** people can be seated.

If we think about how we can seat the kids so that the boys won’t sit next to each other, we realize that each pair of boys must be separated by a girl. Thus, the only arrangement in terms of gender is **BGBGBGB**. Since we have the arrangement in place, the number of ways to do this depends only on what order the boys sit in, and the order in which the girls sit. This is **4!** for the boys and **3!** for the girls, so the total of **4!*3!**.

Now the total number of ways to sit **7** children is simply **7!**.

Thus our final answer becomes **7!-4!*3! = 4896**.

**Example 10:**

**Q:** Find the total number of ways in which **10** distinct objects can be put into **2** different boxes so that no box remains empty?

**Solution:**

Again, using our complimentary counting skills, we first want to see how many ways we can put these **10** objects in these **2** boxes such that at least one of them remains empty. We can then subtract this from the total number of ways. Both boxes can’t be empty, so either the first is empty or the second is empty. Thus there are only two ways of this happening.

Now we find the total number of ways. Each object has 2 choices: it can be in the first or the second box. Thus the total number is **2*2*2*..*2 = ** **ways**

Therefor, our final answer becomes ** **.

** Exercises:**

1. How many four letter code words are possible using first 10 letters of the English alphabet, if no letter can be repeated?

2. How many numbers are there between 100 and 1000 such that every digit is either 2 or 9?

3. How many committees of seven with a given chairman can be selected from twenty people?

4. How many hand shakes if 10 friends meet and shake hands with one another?

5. In how may ways a deck of cards be arranged if no two hearts are adjacent?

6. Three distinguishable dice are thrown. In how many ways they can land and give a sum of 9?

7. In how many ways one permute the word MEPHISTOPHELES?

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# Cube Coloring

Suppose we have a 3X3X3 cube. Dip the cube in red paint and split it into 27 cubes each with size 1X1X1. How many of the 1X1X1 cubes are painted on exactly two faces?

A cube has 8 vertices, 12 edges and 6 faces. Since three faces meet at each of the 8 vertices in a cube, the eight 1X1X1 cubes at each vertex will have 3 faces painted. Cubes with two faces painted are the cubes corresponding to the 12 edges of the 3X3X3 cube. Six of the 1X1X1 cubes in each face of the 3X3X3 are painted one face.

The number of 1X1X1 cubes with:

3 faces painted = 8

2 faces painted = 12

1 face painted = 6

No faces painted=1

So there are **twelve** 1X1X1 cubes with two faces painted.

It is clear that the vertices, edges and the faces of the cube play a major role in finding the painted faces when splitting a 3X3X3 cube into 27 1X1X1 cubes.

Is it possible to generalize this? Yes. For any integer N>= 3, consider the NxNxN cube. Paint this cube and then split it into 1X1X1 cubes. Clearly we will get N^{3} 1X1X1 cubes.

The number of 1X1X1 cubes with:

3 faces painted = 8

2 faces painted =

1 face painted =

No faces painted=

We can see that

We will look at example now. Take N=5. Then the cube is 5X5x5. After painting, we will get 125 1X1X1 cubes.

The number of 1X1X1 cubes with:

3 faces painted = 8

2 faces painted = = 12(5-2)=36

1 face painted = = 6(5-2)^{2}=54

No faces painted= = (5-2)^{3} = 27.

We will look at some problems based on this concept.

- In a birthday party, an NxNxN cubical special cake frosted on all 6 faces was cut into 1X1X1 cakes and distributed to all the participants. The number of people who got the cakes with no frost in it was 8 times the number of people who received the cakes with 3 faces frosted. How many people were there in the party?

Solution: Since there are 8 cakes with 3 faces frosted(the vertices), the number of people who got cakes with no frost is 8X8 = 64.

So, = 64 = 4^{3}

So, N=6.

Therefore: NxNxN = 6X6X6 = 216

Number of people in the party is 216.

Try these problems:

2. A cube of cheese is 4 cm wide, 4 cm long and 4 cm high. Three faces of the cube that meet in a corner are covered with thin layer of wax. The cheese is then cut into 64 small cubes with sides of length 1 cm. How many of these small cubes have no wax on them?

3. A cubical box without a top is 4 cm on each edge. It contains 64 identical 1-cm cubes that exactly fill the box. How many of these small cubes actually touch the box?

# Amicable numbers

The unravelling of the hidden properties of numbers never fails to fascinate math enthusiasts. Pythagoreans, known for the famous Pythagoras Theorem, saw a property for the pair of numbers 220 and 284 which they called Amicable. It started the hunt for other pairs of amicable numbers. What is that make pair of numbers amicable or friendly?

Find the proper divisors (all divisors except the number itself) of 220 and add them. Guess what? You will get 284. If you do the same with 284, you will end up with 220.

The sum of proper divisors of 220:

1+2+4+5+10+11+20+22+44+55+110 = 284

The sum of the proper divisors of 284:

1+2+4+71+142 = 220.

Two positive integers are amicable if the sum of proper divisors of one integer equals the other.

After Greeks, Arab mathematicians found several amicable numbers. Euler has given a rule to produce amicable numbers. We will look at Euler’s rule. Take prime numbers p, q, r as given below with n and m as positive integers with n > m.

Then are an amicable pair of integers.

For example, take n=2 and m=1. Then , q= 11 and r = 71. It follows that 4*5*11= 220 and 4*71= 284, the same two numbers we explored at the beginning. Euler foundnearly 60 amicable pairs using this rule.

However, Euler’s rule doesn’t provide all possible amicable numbers. There are amicable numbers not satisfying this rule. The amicable pair (1184, 1210) found by a sixteen year old Italian boy Nicolo Paganini in 1866 doesn’t satisfy Euler’s rule.

Please look at this nice video presented by Dr. James Grime on Amicable numbers.

Also look at http://mathworld.wolfram.com/AmicablePair.html for further reading.

# 2014

Happy New Year! I start this new blog by posting some interesting facts about the number 2014. This post is mainly addressed to young students. 2014 is an even number. Let us see whether it is **a perfect number**. What is a perfect number?

If the sum of the positive proper divisors of a positive number equals the number, then it is called a perfect number. Here are some examples of perfect numbers

**6=1+2+3
28=1+2+4+7+14
496=1+2+4+8+16+31+62+124+248**

In the case of 2014,

2014 = 2X19X53.

So there are 8 factors for 2014. When you sum the proper factors of 2014, we get

1+2+19+53+38+106+1007 = 1226.

Since 1226 is less than 2014, 2014 is a deficient number.

The decimal system is the one that we use in day to day life. Roughly speaking, the digits 0,1,2,3,4,5,6,7,8,9 are used to express any number and perform arithmetic operations. For example, 123 has one hundred, two tens and three ones.

Rather, 123=1X10^2+2X10+3X1

Why should not we use the two digits 0 and 1 to express the numbers instead of ten digits?

Yes. We can. It is called the binary number system. The number 5 can be expressed as 101 in the binary form.

101 = 1X2^2+0X2+1X1=4+0+1=5.

If a number, expressed in binary form, contains odd number of 1s, then the number is called as an **Odious number**. On the other hand, if it contains even number of 1s, it is called as an **Evil number.**

The binary expansion of 2014 is 11111011110. , It has odd number of 1s. So 2014 is an Odious number

Even though 2014 is not a prime number, the sum of the digits and the sum of the cubes of the digits of 2014 are prime.

2+0+1+4=7 and 8+1+64=73.

Using the four symbols plus, minus, multiplication and division, 2010 can be expressed as

2010 = (7X 9+4)X 5X (6/2)X (3-1+8X 0).

Can you try expressing 2014 in a similar way?

That is, using the digits 1 2 3 4 5 6 7 8 9 0 once with the four basic arithmetic operations.

We will discuss about perfect numbers in detail in future posts.