Algebraic Expansions

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Pascal's Triangle

Here it is below:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1

Study these numbers and see if you can figure what the next line should be before reading on . . .

Each number in the triangle is the sum of two above. For example, the 6 on line 5 is the sum of the pair of 3's above. So the next line is

1, 10 (1 + 9), 45 (9 + 36), 120 (36 + 84), etc.

I am now apparently changing the subject and turning to a bit of algebra.

Algebraic Expansions

Imagine having to expand an expression like

(1 + x)²

By "expanding" I mean remove the brackets. As a reminder, the expression is expanded like this:

(1 + x)² = (1 + x)(1 + x) = 1 + 2x + x²

In the middle pair of brackets, each term in the left bracket is multiplied by each term in the right bracket. We can do the same with a cube, i.e (1 + x)³.

(1 + x)³ = (1 + x)(1 + x)(1 + x) = (1 + x)(1 + 2x + x²)
= 1 + 3x + 3x² + x³

If you look at the coefficients (the numbers on their own and in front of the x's) of the results you will see that for the first one they are 1, 2, 1 and for the second one they are 1, 3, 3, 1. These, of course, are the lines from Pascal's Triangle. And yes, it does work for all positive whole number values of the index. Prove to yourself by algebra that,

(1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴

In fact there is a general rule that

(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

As you see, the indexes begin at 4 and descend for the a's while they ascend to 4 for the b's. The coefficients are the 1, 4, 6, 4, 1 from Pascal's triangle. So let's try an example.

Expand (2 + 3x)⁵.

Applying the general rule of ascending and descending indexes and the coefficients from Pascal we can immediately expand the above equation (if we set a to 2 and b to 3):

2⁵ + 5(2⁴)(3x) + 10(2³)(3x)² + 10(2²)(3x)³ + 5(2)(3x)⁴ + (3x)⁵

This simplifies to

32 + 240x + 720x² + 1080x³ + 810x⁴ + 243x⁵

It is now time to apparently change the subject again...

Selections and Combinations

Imagine you have five books. Let us imagine that you wanted to read one. How many ways are there of selecting a single book? Well, that is easy, there are 5 ways. If we label the books A, B, C, D, E, we can chose any one of five; five different selections.

How many selections are there if we wanted to select two books? Well, let's list all the combinations:

AB, AC, AD, AE, BC, BD, BE, CD, CE, DE.

That makes 10 selections.

What about if we wanted three from five? Well, that's easy. Picking three from five is the same as discarding two from five so there are ten ways of doing this. Picking four from five is the same as discarding one from five, five ways. Of course, if you wanted to select all five books there is only one way of doing that. There is also just one way of selecting no books! So tabulating we have:

Number of selections from 5	0	1	2	3	4	5
Number of ways of making selections	1	5	10	10	5	1

Again, these numbers are a line from Pascal's Triangle.

This is a fascinating aspect of mathematics that two seemingly unconnected topics are in fact connected. In this case we have a connection between Algebraic expansions and selections.

The number of ways of selecting r objects from a total of n is written as

ⁿC_r

So the above selections can be written mathematically as follows:

There is one way to select no books from 5	⁵C₀ = 1
There are 5 ways of selecting 1 book from 5	⁵C₁ = 5
There are 10 ways of selecting 2 books from 5	⁵C₂ = 10
There are 10 ways of selecting 3 books from 5	⁵C₃ = 10
There are 5 ways of selecting 4 books from 5	⁵C₄ = 5
There is one way to select 5 books from 5	⁵C₅ = 1

Question: "How many ways are there of selecting six objects from eight?". Mathematically, the answer is⁸C₆. From Pascal's Triangle, there are total of eight objects so you look at the line beginning with 1, 8, etc. You want to select six so you count along from zero, until you count six. The number there is 28 so there are 28 selections.

Mathematically we can write ⁸C₆ = 28.

Rather than having to memorise Pascal's Triangle, it would be useful if there was a formula for calculating ⁿC_r. It could be used both for selections and for expanding algebraic expressions. Remember that

(1 + x)² = ²C₀ + ²C₁x + ²C₂x²

= 1 + 2x + x².

There is a formula and it is very easy to use but looks difficult. It may be useful to re-read the first part of this essay again.

Formula for ⁿC_r

The formula for ⁿC_r is:

ⁿC_r = n! / (n - r)! r!

The term n! (pronounced n factorial) means multiply together all the whole numbers from 1 to n. So,

1! = 1

2! = 1 x 2 = 2

3! = 1 x 2 x 3 = 6

4! = 1 x 2 x 3 x 4 = 24

5! = 1 x 2 x 3 x 4 x 5 = 120

Also, 0! is defined as 1.

It is now time to return to selections. How many ways can 8 books be selected from 11? The answer is, of course ¹¹C₈ which is given by:

¹¹C₈ = 11! / (11 - 8)! 8! = 11! / 3! 8!
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 / 1 x 2 x 3 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8
= 9 x 10 x 11 / 1 x 2 x 3

This is now a case of cancelling down to 165. So there are 165 ways of selecting 8 books from 11.

Using the above formula instead of Pascal's Triangle we can look at one of our expansions and say that

(1 + x)³ = ³C₀ + ³C₁x + ³C₂x² + ³C₃x³

and show that

³C₀ = 3! / 3! 0! = 1 x 2 x 3 / 1 x 2 x 3 x 1 = 1 (remember 0! = 1)

³C₁ = 3! / 2! 1! = 1 x 2 x 3 / 1 x 2 x 1 = 3

³C₂ = 3! / 1! 2! = 1 x 2 x 3 / 1 x 1 x 2 = 3

³C₃ = 3! / 0! 3! = 1 x 2 x 3 / 1 x 1 x 2 x 3 = 1.

Thus the numbers of Pascal's Triangle can be calculated from the formula above for ⁿC_r. We can do this for any value of n that is a positive whole number.

In summary, Pascal's Triangle can be used:

to form the basis of selection probabilities (these are called Combinations, hence the C in the ⁿC_r above).
to determine the coefficients of the expansions of (1 + x)ⁿ, where n is a positive whole number.

Just to remind you, when n is a positive whole number the expansion contains x + 1 terms.

The expansion for (1 + x)³ contains 4 terms.

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Algebraic Expansions

Selections and Combinations

Formula for nCr

Formula for ⁿC_r