The Binomial Theorem by David Grisman Introduction The binomial theorem is used to evaluate the term (a+b)n. To understand why this is necessary, let us make an attempt to evaluate (a+b)n using the current method of distribution (also known as FOILing). For n = 2, we obtain:

(a + b )2 = (a + b )(a + b ) = a 2 + ab + ab + b 2 = a 2 + 2ab + b 2 Likewise, we could do the same for n = 3: (a + b )3 = (a + b )2 (a + b )

(

)

= a 2 + 2ab + b 2 (a + b ) = a 3 + a 2 b + 2a 2 b + 2ab 2 + ab 2 + b 3 = a 3 + 3a 2 b + 3ab 2 + b 3 We can see, therefore, that as n increases, evaluation by distribution becomes more and more complex. Finding (a+b)12, for example, would be extremely tedious. We would much rather have a formula in terms of n so that we can evaluate (a+b)n by simply plugging in our value of n. This is the binomial theorem:

( n % n!k k (a + b ) = " && ##a b k =0 ' k $ n

n

Because there may be some mathematical symbols in the above equation that seem unfamiliar, this document is designed to walk through each step of the binomial theorem, briefly explain some of its implications, and give examples of its applications.

Sigma Notation Though sigma notation is probably a familiar concept, a brief review will be helpful. Sigma notation is used to represent a series – a group of numbers added together. For example: 5

2

! (3k + 1) k =0

This particular series has six terms, each corresponding to a given value of k. The number below the ∑ is where the sum begins and the number above is where it ends. For this example, the terms correspond to k = 0, 1, 2, 3, 4, 5. 2 Term 1 (k = 0): (3 ! 0 + 1) = 1 2 Term 2 (k = 1): (3 !1 + 1) = 16 2 Term 3 (k = 2): (3 ! 2 + 1) = 49 2 Term 4 (k = 3): (3 ! 3 + 1) = 100 2 Term 5 (k = 4): (3 ! 4 + 1) = 169 2 Term 6 (k = 5): (3 ! 5 + 1) = 256

To evaluate the series, we simply add the terms together: 5

2

! (3k + 1)

= 1 + 16 + 49 + 100 + 169 + 256

k =0

= 591

Sigma notation will be used heavily throughout this document.

Factorials Factorials are probably the most exciting part of mathematics because we use exclamation points (!) to denote them. Factorials may look confusing but are actually fairly simple. In the general case, n! (read as “n factorial”) can be evaluated as n!= n ! (n " 1)! (n " 2 )! ! ! 3 ! 2 ! 1 For example, let’s evaluate 5!: 5!= 5 ! 4 ! 3 ! 2 ! 1 = 120

Now let’s evaluate 10!: 10!= 10 ! 9 ! 8 ! 7 ! 6 ! 5 ! 4 ! 3 ! 2 ! 1 = 3628800

Also, by definition, 1! = 1 and 0! = 1. We will use factorials to define new terms in our next section.

Permutations Let us consider an apartment of six burly men: Peter, Dallas, Dustin, David, Robert, and Daniel. This apartment has six beds, numbered 1-6. None has chosen which bed he wants to sleep in yet, and each must choose a separate bed. How many different ways are there to order these six burly men so that each gets a different bed? Another way to word the question is: how many permutations are there of these six men? To the right is a diagram showing the different possibilities. However, it is important to note that once Peter has chosen a bed (say, for example, Bed 1), that bed can no longer be chosen by any other roommate. Also, Peter could no longer choose another bed. To find the total number of possibilities, we start with the first bed. Bed 1 has six possible owners. Once it has been chosen, though, Bed 2 now only has five possible owners. Likewise, once Bed 2 has been chosen, Bed 3 only has four possible owners, and so forth. So, Total Possibilities = POBed1· POBed2 · POBed3 · POBed4 · POBed5 · POBed6, where POBedN is # of possible owners for Bed N = 6 ! 5 ! 4 ! 3 ! 2 !1 = 6!

Therefore, we could just obtain the total possibilities, or total permutations, by taking the factorial of the number of burly men. Let’s look at another example: Question: How many ways are there to arrange letters A-Z so that each letter is used exactly once? Solution: There are 26 letters available and 26 positions in which to put them. The first position has 26 possible letters that could fill it. However, once a letter is chosen to fill the first position, there are only 25 possible letters that could fill the second position. Again, once we have chosen a letter for the second position, there are only 24 possible letters for the third position. To obtain the total permutations, we simply multiply them together: Total Permutations = 26! = 26·25·24···2·1 ≈ 4.03·1026

n choose i Now let us take what we have learned from permutations and apply it to a different case. Let’s say, for example, that we have 5 marbles: 2 black and 3 white. If we were to arrange them in different orders, how many different arrangements could we create? One might expect that the answer would be 5! = 120 different arrangements. However, in the above discussion of permutations, we assumed that each object chosen was unique. In the alphabet example, each letter is different from the rest. A is distinguishable from B, B is distinguishable from C, and so forth. Now, though, we don’t have 5 distinct marbles – we only know 2 are black and 3 are white. The total number of arrangements is shown below: BBWWW BWBWW WBBWW BWWBW WBWBW WWBBW WWBWB WWWBB WBWWB BWWWB Obviously, there are not 120 – there are only 10. To resolve this, we must recognize that there would be 120 permutations if we could tell the black marbles from each other and the white marbles from each other. For a

moment let’s assume that we can. We’ll label our marbles W1, W2, W3, B1, and B2. We know that # permutations for B1, B2 = 2! =2 # permutations for W1, W2, W3 = 3! =6 Therefore, to obtain our desired result, we simply take the number of permutations in the distinct case and divide out both the number of permutations among the black and the number of permutations among the white.

120 2!6 = 10 This result matches our previous one. To generalize this more, let i = # of white marbles, and let n = # of total marbles. Then we can see that # non-distinct permutations =

# non-distinct permutations =

&n# n! = $$ !! , i!(n ' i )! % i "

&n# where $$ !! is read as n choose i. Let’s try another example. Suppose we %i " have 10 marbles: 3 black and 7 white. How many arrangements can we make? In this case, i = 7 and n = 10 &n# # arrangements = $$ !! %i " &10 # = $$ !! %7 " 10! = 7!(10 ! 7)! 10 ! 9 ! 8 ! 7 ! 6 ! 5 ! 4 ! 3 ! 2 ! 1 = (7 ! 6 ! 5 ! 4 ! 3 ! 2 ! 1) ! (3 ! 2 ! 1) 10 ! 9 ! 8 = 3 ! 2 !1

= 120 Now, instead of letting i = # of white marbles, we’ll let i = # of black marbles.

&10 # = # of arrangements $$ !! %3 " 10! = 3!(10 ! 3)! = 120 As we’ve shown here, we can let i be either the # of black marbles or the # of white marbles. Both ways will produce the same result. This conclusion leads us to the property

&n# & n # $$ !! = $$ !! . i n ' i % " % " Two additional important properties are

&n# $$ !! = 1 %0" &n# $$ !! = n 2. %1 " Using n choose i, we can determine non-distinct permutations. 1.

The Binomial Theorem Now that we have analyzed all the necessary tools, we are ready to look at the binomial theorem. Recall from the introduction that (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3

= a 3 b 0 + 3a 2 b + 3ab 2 + a 0 b 3

Notice that the exponent of a + the exponent of b = 3 in each term and that each possible exponent of a and b is represented. In general, when we evaluate (a+b)n, we find that

(a + b )n

= c0 a n b 0 + c1 a n !1b1 + c 2 a n ! 2 b 2 + ... + c n !1 a 1b n !1 + c n a 0 b n ,

where c0, c1, c2, … cn-1, cn are coefficients to the ab terms. Now that we can find all the ab terms given any value of n, how, then, could we find the values of all the coefficients? Let’s take a look for the case where n = 5:

(a + b )5 = (a + b )3 (a + b )2 = (a 3 + 3a 2 b + 3ab 2 + b 3 )(a 2 + 2ab + b 2 )

= a 5 + 2a 4 b + a 3b 2 + 3a 4 b + 6a 3b 2 + 3a 2 b 3 + 3a 3b 2 + 6a 2 b 3 + 3ab 4 + a 2 b 3 + 2ab 4 + b 5 = a 5 + 5a 4 b + 10a 3 b 2 + 10a 2 b 3 + 5ab 4 + b 5 In this example, c0 = 1, c1 = 5, c2 = 10, c3 = 10, c4 = 5, and c5 = 1. If we look closely at the coefficients, we can see a trend. Take, for instance, 10a3b2. This term can be rewritten as 10aaabb. How many unique ways could we order aaabb? In other words, how many ways can we choose three a’s and two b’s? &5# # of ways = $$ !! % 2" 5! = 2!!3! 5 ! 4 ! 3 ! 2 !1 = (2 ! 1) ! (3 ! 2 ! 1) 5!4 = 2 !1 = 10 Notice that this corresponds with the coefficient of a3b2. Let’s try it for the term 5ab4. Again, we can rewrite the term: 5abbbb. Does the coefficient correspond to the number of ways we can rearrange abbbb? Let’s find out:

&5# # of ways = $$ !! % 4" 5! = 4!!1! 5 ! 4 ! 3 ! 2 !1 = (4 ! 3 ! 2 ! 1) ! (1) 5 = 1 =5 Our results are again the same. In general, therefore, we can find the coefficients c0, c1, c2, … cn-1, cn by examining the number of ways to rearrange the a’s and b’s in any given term. As we have seen from the above example, we can define the k-th coefficient, ck, to be n choose # of a’s, or n choose # of b’s:

& n # & n # !! = $$ !! ck = $$ % # a ' s " % # b' s " As we examine the equation

we can see that the k-th term also corresponds to the number of b’s in the term (k = 0 term has b0, k = 1 term has b1, k = 2 term has b2, etc.). Therefore, we could simply rewrite our equation for ck:

&n# c k = $$ !! %k " Now let us go back to our general equation for (a+b)n:

We can now plug in our new formula for ck:

(a + b )n

&n# &n# &n# & n # 1 n '1 & n # 0 n !!a b + $$ !!a b = $$ !!a n b 0 + $$ !!a n '1b1 + $$ !!a n ' 2 b 2 + ... + $$ %0" %1 " %2" % n ' 1" %n"

Rewriting using sigma notation, we now obtain:

(a n+ b )nn =(cn0 a% nbn!0k+ ck 1a n!1b1 + c2 a n!2b 2 + ... + cn!1a1b n!1 + cn a 0b n (a + b ) = " && ##a b , k =0

'k $ k=0 term

k=1 term

k=2 term

which is also called the binomial theorem. We now have a generalized formula for finding (a+b)n. Let’s use it to evaluate (a+b)4:

k = n-1 term

4 % 4" k k ##a b k k =0 ' $ 4

(a + b )4 = ! (&&

& 4# & 4# & 4# & 4# & 4# = $$ !!a 4 b 0 + $$ !!a 3 b1 + $$ !!a 2 b 2 + $$ !!a 1b 3 + $$ !!a 0 b 4 %0" %1 " % 2" %3" % 4" 4! 4! = 1 ! a 4 + 4 ! a 3b + ! a 2b 2 + ! ab 3 + 1 ! b 4 2!!2! 3!!2! 4 ! 3 ! 2 !1 2 2 4 ! 3 ! 2 !1 = a 4 + 4a 3 b + a b + ab 3 + b 4 (2 ! 1)! (2 ! 1) (3 ! 2 ! 1) ! (1) = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4 To check the answer, we can use distribution:

(a + b )4 = (a + b )2 (a + b )2

(

)(

= a 2 + 2ab + b 2 a 2 + 2ab + b 2

)

= a 4 + 2a 3 b + a 2 b 2 + 2a 3 b + 4a 2 b 2 + 2ab 3 + a 2 b 2 + 2ab 3 + b 4 = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4

k=n term

This result is the same as before, validating our answer. In general, distribution is a really messy process, so using the binomial theorem will suffice.

Pascal’s Triangle The binomial theorem in its modern form is attributed to Blaise Pascal, who first published it in a pamphlet in 1665. Later he developed a form of the binomial theorem that made the information easier to access. This form is known as Pascal’s Triangle. To see how it is set up, examine (a+b)n for n = 0 through n = 5:

(a + b )0 = 1 (a + b )1 = a + b (a + b )2 = a 2 + 2ab + b 2 (a + b )3 = a 3 + 3a 2 b + 3ab 2 + b 3 (a + b )4 = a 4 + 4a 3b + 6a 2 b 2 + 4ab 3 + b 4 (a + b )5 = a 5 + 5a 4 b + 10a 3b 2 + 10a 2 b 3 + 5ab 4 + b 5 Already we see the beginnings of a triangle. If we take away the left sides of each of the above equations and only use the coefficients of the right sides, we obtain Pascal’s Triangle:

Each row on the Triangle corresponds to a successive value of n (n = 0 for row 1, n = 1 for row 2, n = 2 for row 3, etc.). Initially, the value of the Triangle may not be obvious. Of course, if we had to go through the binomial theorem to obtain the values for the Triangle, the Triangle would

be useless. If we look at the Triangle a little closer, though, we can find an interesting property: any value can be calculated by simply adding the two numbers directly above it.

This property of Pascal’s Triangle allows us to be able to easily calculate new rows as needed. For instance, let’s evaluate (a+b)7:

From our new rows on the triangle, we get:

(a + b )7

= a 7 + 7 a 6 b + 21a 5 b 2 + 35a 4 b 3 + 35a 3 b 4 + 21a 2 b 5 + 7 ab 6 + b 7

Pascal’s Triangle gives us the opportunity to evaluate (a+b)n without having to multiply or distribute.

Appendix A: Example Problems Sigma Notation

k2 +3 Problem: Evaluate ! . k = "4 12 3

Solution:

(!4) 2 + 3 19 = Term 1 (k = -4): 12 12 2 (!3) + 3 12 = Term 2 (k = -3): 12 12 2 (!2) + 3 7 = Term 3 (k = -2): 12 12 2 (!1) + 3 4 = Term 4 (k = -1): 12 12 02 + 3 3 = Term 5 (k = 0): 12 12 2 1 +3 4 = Term 6 (k = 1): 12 12 2 2 +3 7 = Term 7 (k = 2): 12 12 2 3 + 3 12 = Term 8 (k = 3): 12 12 2 3 k + 3 19 12 7 4 3 4 7 12 = + + + + + + + ! 12 12 12 12 12 12 12 12 12 k = "4 68 = 12 Factorials Problem: Evaluate 13! Solution: 13! = 13·12·11·10·9·8·7·6·5·4·3·2·1 ≈ 6.227·109

Permutations Problem: How many different combinations of four letters can be made if none of the letters can be repeated? Solution: To begin, there are 26 letters in the alphabet. The first letter can be any of the 26. The second can be any letter except that which has already been chosen, making only 25 possibilities. The third and the fourth only have 24 and 23 possibilities, respectively. To obtain the total number of combinations, we simply multiply these together: # permutations = 26·25·24·23 = 358,800 n choose i Problem: A student committee is to be chosen from among 30 candidates, 13 of whom are boys. The committee must have 6 people: 3 boys and 3 girls. How many different committees can be formed? Solution:

&13 # # combinations of boys = $$ !! %3 " 13! = 10!!3! 13 ! 12 ! 11 ! 10! = 10!!3! 13 ! 12 ! 11 = 3 ! 2 !1 = 286 &17 # # combinations of girls = $$ !! %3 " 17 ! 16 ! 15 ! 14! = 14!!3! 17 ! 16 ! 15 = 3 ! 2 !1

= 680 Total combinations = # combinations of boys · # combinations of girls = 286 · 680 = 194,480 Binomial Theorem Problem: Evaluate (a+b)6 using the binomial theorem. Solution:

6 % 6" k k ##a b k k =0 ' $ 6

(a + b )6 = ! (&&

& 6# & 6# &6# & 6# &6# & 6# = $$ !!a 6 b 0 + $$ !!a 5b1 + $$ !!a 4 b 2 + $$ !!a 3b 3 + $$ !!a 2 b 4 + $$ !!a1b 5 % 0" %1 " % 2" %3" % 4" %5" & 6# + $$ !!a 0 b 6 % 6" = a 6 + 6a 5 b + 15a 4 b 2 + 20a 3 b 3 + 15a 2 b 4 + 6ab 5 + b 6 Pascal’s Triangle Problem: Evaluate (a+b)6 using Pascal’s Triangle. Solution: The row corresponding to n = 6 is: 1

6

15

20

15

6

1

as is shown in the text. To find (a+b)6, we simply use the coefficients given: (a + b) 6 = a 6 + 6a 5b + 15a 4 b 2 + 20a 3b 3 + 15a 2 b 4 + 6ab 5 + b 6 This matches the result obtained in the previous problem.