1.5 Binomial Theorem & Counting

The binomial expansion of `(1 + kx) ^ n` is given by `1 + (9x)/2 + 15k^2 x^2 + ... + k^n x^n` where `n in Z^+ and k in Q`

Find the value of n and the value of k .

ⁿC₂ = 15

`nk= 9/2`

`(n(n-1))/2=15`

`n(n-1)=30`

finding correct value in Pascal's triangle

`=> n = 6 and k = 3/4`

Consider the expansion of `(ax + 1)^9 / (21x^2), " where " a != 0`.The coefficient of the term in `x^4`is `8/7a^5`

Find the value of a.

correct term in `x^6`or or coefficient of `x^6`

(may be seen in equation)

`84a^6x^6`

set their term in `x^6` or coefficient of `x^6` equal to `24a^5x^6` or 24`a^5` (do not accept other powers of x)

`8a = 24`

THEN a = `2/7 or 0.286(0.285714...)`

Consider the binomial expansion `(x + 1)^7 = x^7 + ax^6 + bx^5 + 35x^4 + ... + 1`, `"where " x != 0 " and " a, b in ZZ^+`

(a) Show that `b = 21` . The third term in the expansion is the mean of the second term and the fourth term in the expansion.

(b) Find the possible values of `x`.

(a) lists terms from row 7 of Pascal's triangle `1,7,21,...`

THEN b = 21

(b) correct equation `21x^5 = (ax^6 + 35x^4) / 2`

correct quadratic equation `7x^2 - 42x + 35 = 0`

valid attempt to solve their quadratic `(x - 1) (x -5) = 0`

`=> x = 1; x = 5`

Consider the expansion of`(3 + x^2)^(n+1)`, where `n in Z^+`. Given that the coefficient of `x^4` is 20412, find the value of n.

`3^(n+1) (1 + x^2 / 3)^(n+1)`

product of a binomial coefficient, and a power of `x^2/3` OR `1/3`  seen evidence of correct term chosen

`3^(n+1) xx (n(n+1))/(2!) xx (x^2 / 3)^2 (= (3^(n-1) n(n+1))/2 x^4)`

equating their coefficient to `20412` or `20412x^4` their term to  

`3^(n-1) xx (n(n+1))/2 = 20412`

`n = 7`

In the expansion of `(x + k)^7`, where `k in R` the coefficient of the term in `x^5` is 63. Find the possible values of k.

attempt to use the binomial expansion of `(x+k)^7`

⁷C₀`x^7k^0`+ ⁷C₁`x^7k^1` + ⁷C₂`x^7k^2`

identifying the correct term ⁷C₂`k^2x^5`

A farmer has six sheep pens, arranged in a grid with three rows and two columns as shown in the following diagram.

Five sheep called Amber, Brownie, Curly, Daisy and Eden are to be placed in the pens. Each pen is large enough to hold all of the sheep. Amber and Brownie are known to fight.

Find the number of ways of placing the sheep in the pens in each of the following cases:

(a)  Each pen is large enough to contain five sheep. Amber and Brownie must not be placed in the same pen

(b) Each pen may only contain one sheep. Amber and Brownie must not be placed in pens which share a boundary.

(a)

total number of ways  = `6^5`

number of ways with Amber and Brownie together  = `6^4`

attempt to subtract (may be seen in words)  = `6^5 - 6^4 = 6480`

(b)

total number of ways  `= 6! = 720`

number of ways with Amber and Brownie sharing a boundary  = 2 x 7 x 4! = 336

attempt to subtract (may be seen in words) = 720 - 336 = 384

Find the term independent of `x` in the expansion of `1/x^3 (1/(3x^2) - x/2)^9`

use of Binomial expansion to find a term in either

`(1/(3x^2) - x/2)^9 , (1/3 x^-2 - 1/2 x)^9 , (1/3 - x^3 / 2)^9 , (1/(3x^2) - x/2)^9 , (1/3 x^-2 - 1/2 x)^9 " or " (2 - 3x^3)^9`

finding the powers required to be 2 and 7 

constant term is `(9/2)`x `(1/3)^2`x `(-1/2)^7`

therefore term independent of `x` is `-1/32 =-0.03125`

Mary, three female friends, and her brother, Peter, attend the theatre. In the theatre there is a row of 10 empty seats. For the first half of the show, they decide to sit next to each other in this row.

(a) Find the number of ways these five people can be seated in this row.

For the second half of the show, they return to the same row of 10 empty seats. The four girls decide to sit at least one seat apart from Peter. The four girls do not have to sit next to each other.

(b) Find the number of ways these five people can now be seated in this row.

(a)

6 x 5! =720 (accept 6!)

(b)

Peter apart from girls, in an end seat = ⁸P₄ = 1680 OR 

Peter apart from girls, not in end seat = ⁷P₄ = 840

case 1: Peter at either end

2 x ⁸P₄ = 2 x 1680 = 3360

case 2: Peter not at the end

8 x ⁷P₄ = 8 x 840 = 6720 

Total number of ways = 6720 + 3360 = 10080

Consider the expression `1 / sqrt(1 + ax) - sqrt(1 - x) " where " a in QQ, a != 0`

The binomial expansion of this expression, in ascending powers of `x`, as far as the term in `x^2` is `4bx + bx^2, " where " b in QQ`

(a) Find the value of a and the value of b.

(b) State the restriction which must be placed on `x` for this expansion to be valid.

(a)

attempt to expand binomial with negative fractional power

`1 / sqrt(1 + ax) = (1 + ax)^(-1/2) = 1 - (ax)/2 + (3a^2 x^2)/8 + ...`

`sqrt(1 - x) = (1 - x)^(1/2) = 1 - x/2 - x^2/8 + ...`

`1 / sqrt(1 + ax) - sqrt(1 - x) = (1 - a)/2 x + (3a^2 + 1)/8 x^2 + ...`

attempt to equate coefficients of `x` or  `x^2`

`x: (1 - a) / 2 = 4b; x^2: (3a^2 + 1) / 8 = b`

attempt to solve simultaneously

`a = -1/3; b = 1/6`

(b)

|`x`| < 1

Eight runners compete in a race where there are no tied finishes. Andrea and Jack are two of the eight competitors in this race.

Find the total number of possible ways in which the eight runners can finish if Jack finishes.

(a) in the position immediately after Andrea;

(b) in any position after Andrea.

(a)

Jack and Andrea finish in that order (as a unit) so we are considering the arrangement of 7 objects.

7! = 5040 (ways)

b)

the number of ways that Andrea finishes in front of Jack is equal to the number of ways that Jack finishes in front of Andrea

total number of ways is 8!

 `(8!)/2`​ = 20160 (ways)