So Pascal's triangle (as we have just seen) has connections to counting techniques and now it is time to formalise these connections.
Our last activity was this one...
Write down the following (on one line like I have)
Then underneath that write the answers....
What connection do you see to Pascal's triangle?
What was the connection?
You should have discovered that the above answers to the combinations was an entire row from Pascal's triangle.
The $n$n value is the row number (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 ^{9}C_{4}, will be the row beginning 1--9..... and be the $5$5^{th} number in the row - (remember we start the element from $0$0).
Find the missing elements in the 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 ^{9}C_{4} and $\editable{C}$C =^{9}C_{5}, but we also know that ^{9}C_{4}=^{9}C_{5} (confirming what we already knew from symmetry that the values will be the same).
$\editable{B}$B $=$= ^{9}C_{4} $=126$=126
Thus both $\editable{B}$B and $\editable{C}=126$C=126.
$(a+b)^0$(a+b)0 | $=$= | $1$1 |
$(a+b)^1$(a+b)1 | $=$= | $a+b$a+b |
$(a+b)^2$(a+b)2 | $=$= | $a^2+2ab+b^2$a2+2ab+b2 |
$(a+b)^3$(a+b)3 | $=$= | $a^3+3a^2b+3ab^2+b^3$a3+3a2b+3ab2+b3 |
$(a+b)^4$(a+b)4 | $=$= | $a^4+4a^3b+6a^2b^2+4ab^3+b^4$a4+4a3b+6a2b2+4ab3+b4 |
$(a+b)^5$(a+b)5 | $=$= | $a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$a5+5a4b+10a3b2+10a2b3+5ab4+b5 |
Consider the expansions above of $(a+b)^n$(a+b)n. Particularly note the following patterns.
What are the coefficients for the expansion of $(a+1)^7$(a+1)7, and then write out the full expansion.
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 expansion of $(a+1)^7$(a+1)7 will be
$\left(a+1\right)^7$(a+1)7 | |
$=$= | $a^7+7a^61^1+21a^51^2+35a^41^3+35a^31^4+21a^21^5+7a^11^6+1^7$a7+7a611+21a512+35a413+35a314+21a215+7a116+17 |
$=$= | $a^7+7a^6+21a^5+35a^4+35a^3+21a^2+7a+1$a7+7a6+21a5+35a4+35a3+21a2+7a+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
How many terms are there in the expansion of $\left(m+y\right)^8$(m+y)8?
Using the relevant row of Pascal’s triangle, determine the coefficient of each term in the expansion 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.
Use permutations and combinations