In general, we can find the values of $\nCr{n}{r}$nCr in row number $n$n (starting at row $0$0) and the $r$r value is the element in the row, (also starting at $0$0).

So the value for $\nCr{9}{4}$9C4, will be the row beginning $1$1, $9$9, ..... and be the $5$5^th number in the row - (remember we start the element from $0$0).

Worked example

Example 1

Find the missing elements in this row from Pascal's Triangle.

$1,9,$1,9, $\editable{A},84,\editable{B},\editable{C},84,36,9,1$A,84,B,C,84,36,9,1

Firstly we know that the lines of the triangle are symmetrical. This helps us identify that box $\editable{A}$Ashould be the value of $36$36. As reading from left to right is the same as reading from right to left.

This symmetry doesn't help us with the values for $\editable{B}$B or $\editable{C}$C, but we can use our knowledge of combinations to solve this.

$\editable{B}=\editable{C}$B=C because of of the symmetry.

$\editable{B}$B also equals the value of $\nCr{9}{4}$9C4 and $\editable{C}$C$=$=$\nCr{9}{5}$9C5, but we also know that $\nCr{9}{4}=\nCr{9}{5}$9C4=9C5 (confirming what we already knew from symmetry that the values will be the same).

$\editable{B}$B $=$= $\nCr{9}{4}$9C4 $=126$=126

Thus both $\editable{B}$B and $\editable{C}=126$C=126.

Pascal's triangle and binomial distributions

$(a+b)^0$(`a`+`b`)0	$=$=	$1$1
$(a+b)^1$(`a`+`b`)1	$=$=	$a+b$`a`+`b`
$(a+b)^2$(`a`+`b`)2	$=$=	$a^2+2ab+b^2$`a`2+2`ab`+`b`2
$(a+b)^3$(`a`+`b`)3	$=$=	$a^3+3a^2b+3ab^2+b^3$`a`3+3`a`2`b`+3`ab`2+`b`3
$(a+b)^4$(`a`+`b`)4	$=$=	$a^4+4a^3b+6a^2b^2+4ab^3+b^4$`a`4+4`a`3`b`+6`a`2`b`2+4`ab`3+`b`4
$(a+b)^5$(`a`+`b`)5	$=$=	$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$`a`5+5`a`4`b`+10`a`3`b`2+10`a`2`b`3+5`ab`4+`b`5

Consider the distributions above of $(a+b)^n$(a+b)n. Particularly note the following patterns.

For each distribution to the power $n$n, there are $n+1$n+1 elements.
For each term, the sum of the exponents is $n$n.
Powers of $a$a decrease from left to right, from $n$n down to $0$0.
Powers of $b$b increase from left to right, from $0$0 up to $n$n.
The coefficients start at $1$1, end at $1$1, and are the terms of the relevant row from Pascals triangle.

Worked example

Example 2

What are the coefficients for the distribution of $(a+1)^7$(a+1)7, and then write out the full distribution.

So we can see that we will have $n+1=8$n+1=8 terms.

We can refer to the relevant row in Pascal's triangle, specifically this row

This shows us that the coefficients will be

$1,7,21,35,35,21,7,1$1,7,21,35,35,21,7,1.

Thus the full distribution of $(a+1)^7$(a+1)7 will be

	$\left(a+1\right)^7$(`a`+1)7
$=$=	$a^7+7a^61^1+21a^51^2+35a^41^3+35a^31^4+21a^21^5+7a^11^6+1^7$`a`7+7`a`611+21`a`512+35`a`413+35`a`314+21`a`215+7`a`116+17
$=$=	$a^7+7a^6+21a^5+35a^4+35a^3+21a^2+7a+1$`a`7+7`a`6+21`a`5+35`a`4+35`a`3+21`a`2+7`a`+1

The binomial theorem

We can concisely summarize the pattern in distributions we have observed as a formula called the binomial theorem. This formula will also allow us to find particular terms in an distribution.

Using our knowledge that for an distribution of $\left(a+b\right)^n$(a+b)n the coefficients will be dictated by the combinations of $\nCr{n}{0}$nC0, $\nCr{n}{1}$nC1, $\nCr{n}{2}$nC2, $\dots$…, $\nCr{n}{n}$nCn, also notated as $\binom{n}{0}$(n0),$\binom{n}{1}$(n1),$\binom{n}{2}$(n2),$...$...,$\binom{n}{n}$(nn)

This results in the distribution looking like this:

$(a+b)^n=$(a+b)n=$\binom{n}{0}$(n0)$a^n$an$+$+$\binom{n}{1}$(n1)$a^{n-1}b^1+$an−1b1+$\binom{n}{2}$(n2)$a^{n-2}b^2+$an−2b2+$\binom{n}{3}$(n3)$a^{n-3}b^3+...+$an−3b3+...+$\binom{n}{r}$(nr)$a^{n-r}b^r+...+$an−rbr+...+$\binom{n}{n-1}$(nn−1)$a^1b^{n-1}+$a1bn−1+$\binom{n}{n}$(nn)$b^n$bn

Thus any particular term can be found using $\binom{n}{r}$(nr)$a^{\left(n-r\right)}$a(n−r)$b^r$br.

Worked examples

Example 3

Distribute $(2x+3)^5$(2x+3)5.

$(a+b)^n=$(a+b)n=$\binom{n}{0}$(n0)$a^n+$an+$\binom{n}{1}$(n1)$a^{n-1}b^1+$an−1b1+$\binom{n}{2}$(n2)$a^{n-2}b^2+$an−2b2+$\binom{n}{3}$(n3)$a^{n-3}b^3+...+$an−3b3+...+$\binom{n}{r}$(nr)$a^{n-r}b^r+...$an−rbr+...$\binom{n}{n-1}$(nn−1)$a^1b^{n-1}+$a1bn−1+$\binom{n}{n}$(nn)$b^n$bn

$(2x+3)^5=$(2x+3)5=$\binom{5}{0}$(50)$(2x)^5+$(2x)5+$\binom{5}{1}$(51)$(2x)^{5-1}3^1+$(2x)5−131+$\binom{5}{2}$(52)$(2x)^{5-2}3^2+$(2x)5−232+$\binom{5}{3}$(53)$(2x)^{5-3}3^3+$(2x)5−333+$\binom{5}{4}$(54)$(2x)^13^{5-1}+$(2x)135−1+$\binom{5}{5}$(55)$(3)^5$(3)5

$(2x+3)^5=1(2x)^5+5(2x)^43^1+10(2x)^33^2+10(2x)^23^3+5(2x)^13^4+1(3)^5$(2x+3)5=1(2x)5+5(2x)431+10(2x)332+10(2x)233+5(2x)134+1(3)5

$(2x+3)^5=32x^5+15(16x^4)+90(8x^3)+270(4x^2)+810x+243$(2x+3)5=32x5+15(16x4)+90(8x3)+270(4x2)+810x+243

$(2x+3)^5=32x^5+240x^4+720x^3+1080x^2+810x+243$(2x+3)5=32x5+240x4+720x3+1080x2+810x+243

Example 4

What is the seventh term in the distribution of $(m-2n)^{12}$(m−2n)12?

We need to construct the seventh term from this $\binom{n}{r}$(nr)$a^{\left(n-r\right)}$a(n−r)$b^r$br where $n$n is $12$12 and $r$r is $6$6.

The coefficient $\binom{n}{r}$(nr) where $n$n is $12$12 and $r$r is $6$6 is $\binom{12}{6}=924$(126)=924.

The term will have both $m$m and $(2n)$(2n) components. The $m$m component would be $m^{12-6}=m^6$m12−6=m6

The $2n$2n component would be $(2n)^6=64n^6$(2n)6=64n6.

So putting that altogether will give us $924m^6\times64n^6=59136m^6n^6$924m6×64n6=59136m6n6.

Practice questions

QUESTION 1

You are given some of the entries in a particular row of Pascal’s triangle. Fill in the missing entries.

$1$1 , $8$8 , $\editable{}$ , $56$56 , $70$70 , $\editable{}$ , $28$28 , $\editable{}$ , $1$1

QUESTION 2

How many terms are there in the distribution of $\left(m+y\right)^8$(m+y)8?

QUESTION 3

Using the relevant row of Pascal’s triangle, determine the coefficient of each term in the distribution of $\left(5+b\right)^5$(5+b)5.

$\left(5+b\right)^5$(5+b)5$=$=$\editable{}$$\times$×$5^5b^0$55b0$+$+$\editable{}$$\times$×$5^4b^1$54b1$+$+$\editable{}$$\times$×$5^3b^2$53b2$+$+$\editable{}$$\times$×$5^2b^3$52b3$+$+$\editable{}$$\times$×$5^1b^4$51b4$+$+$\editable{}$$\times$×$5^0b^5$50b5.

QUESTION 4

Using the binomial theorem, determine the missing powers in the following distribution.

$\left(4p+3q\right)^3=\nCr{3}{0}\left(4p\right)^{\editable{}}\left(3q\right)^0+\nCr{3}{1}\left(4p\right)^{\editable{}}\left(3q\right)^1+\nCr{3}{2}\left(4p\right)^{\editable{}}\left(3q\right)^2+\nCr{3}{3}\left(4p\right)^{\editable{}}\left(3q\right)^3$(4p+3q)3=3C0(4p)(3q)0+3C1(4p)(3q)1+3C2(4p)(3q)2+3C3(4p)(3q)3

QUESTION 5

Distribute $\left(\sqrt{2}x+\frac{1}{y}\right)^4$(√2x+1y)4.