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Combinatorics, EXT1 A1 EQ-Bank 10

By using the fact that  `(1 + x)^11 = (1 + x)^3(1 + x)^8`, show that
 

`((11),(5)) = ((8),(5)) + ((3),(1))((8),(4)) + ((3),(2))((8),(3)) + ((8),(2))`.   (3 marks)

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`text(See Worked Solutions)`

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`text(General term of)\ \ (1 + x)^11 :`

`T_k = \ ^11C_k · 1^(11-k) · x^k`

`=> \ ^11C_5\ text(is the co-efficient of)\ x^5`
 

`(1 + x)^3 = \ ^3C_0 + \ ^3C_1 x + \ ^3C_2 x^2 + \ ^3C_3 x^3`

`(1 + x)^8 = \ ^8C_0 + \ ^8C_1 x + \ ^8C_2 x^2 + \ ^8C_3 x^3 + \ ^8C_4 x^4 + \ ^8C_5 x^5 + …`

 
`:.\ text(Coefficient of)\ x^5\ text(in)\ \ (1 + x)^3(1 + x)^8 `

`= \ ^3C_0 · \ ^8C_5 + \ ^3C_1 · \ ^8C_4 + \ ^3C_2 · \ ^8C_3 + \ ^3C_3 · \ ^8C_2`

`= \ ^8C_5 + \ ^3C_1 · \ ^8C_4 + \ ^3C_2 · \ ^8C_3 + \ ^8C_2`

 
`text(Equating coefficients:)`

`((11),(5)) = ((8),(5)) + ((3),(1))((8),(4)) + ((3),(2))((8),(3)) + ((8),(2))`

Combinatorics, EXT1 A1 SM-Bank 2

Using `(1 + x)^4(1 + x)^9 = (1 + x)^13`

show that

   `\ ^9C_4 + \ ^4C_1\ ^9C_3 + \ ^4C_2\ ^9C_2 + \ ^4C_3\ ^9C_1 + 1 = \ ^13C_4`   (2 marks)

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`text(See Worked Solutions)`

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`text(Expanding)\ \ (1 + x)^13 :`

`T_k = \ ^13C_k · 1^(13-k) · x^k`

`=> \ ^13C_4\ text(is coefficient of)\ x^4`
 

`(1 + x)^4 = \ ^4C_0 + \ ^4C_1 x + \ ^4C_2 x^2 + \ ^4C_3 x^3 + \ ^4C_4 x^4`

`(1 + x)^9 = \ ^9C_0 + \ ^9C_1 x + \ ^9C_2 x^2 + \ ^9C_3 x^3 + \ ^9C_4 x^4 + …`

 

`:.\ text(Coefficient of)\ x^4\ text(in)\ \ (1 + x)^4(1 + x)^9`

`= \ ^4C_0·\ ^9C_4 + \ ^4C_1·\ ^9C_3 + \ ^4C_2·\ ^9C_2 + \ ^4C_3·\ ^9C_1 + \ ^4C_4·\ ^9C_0`

`= \ ^9C_4 + \ ^4C_1·\ ^9C_3 + \ ^4C_2·\ ^9C_2 + \ ^4C_3·\ ^9C_1 + 1`

`= \ ^13C_4\ \ …\ text(as required)`

Combinatorics, EXT1 2018 HSC 14b

  1. By considering the expansions of `(1 + (1 + x))^n` and `(2 + x)^n,` show that
  2. `((n),(r))((r),(r)) + ((n),(r +1))((r + 1),(r)) + ((n),(r + 2))((r + 2),(r)) +`
  3.           `… + ((n),(n))((n),(r)) = ((n),(r)) 2^(n-r)`.   (3 marks)
  4. There are 23 people who have applied to be selected for a committee of 4 people.
  5. The selection process starts with Selector `A` choosing a group of at least 4 people from the 23 people who applied.
  6. Selector `B` then chooses the 4 people to be on the committee from the group Selector `A` has chosen.
  7. In how many ways could this selection process be carried out?   (2 marks)

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i.    `text(Proof)\ \ text{(See Worked Solutions)}`

ii.   `((23), (4)) 2^19`

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i.    `text(Using binomial expansion:)`

`(2 + x)^n= ((n), (0)) 2^n + ((n), (1)) 2^(n-1) x + ((n), (2)) 2^(n-2) x^2 + … + ((n), (n)) x^n`

`[1 + (1 + x)]^n= ((n), (0)) + ((n), (1)) (1 + x) + ((n), (2)) (1 + x)^2 + … + ((n), (n)) (1 + x)^n`
 

`=>\ text(S)text(ince both expansions are equal, we can equate the)`

`text(the coefficients of)\ x^r.`
 

`text(Coefficient of)\ x^r\ text(in the expansion of)\ \ (2+x)^n :`

♦ Mean mark 46%.

`((n), (r)) 2^(n-r) qquad …\ text{(1)}` 

  
`text(Coefficient of)\ x^r\ text(in the expansion of)\ \ [1 + (1 + x)]^n :`

`=>x^r\ \ text(exists in all terms where)\ \ n>=r`

 
`text(Consider the co-efficients of)\ \ x^r ,`

`text(When)\ \ n=r:\ \ ((n), (r))((r), (r))`

`text(When)\ \ n=r+1:\ \ ((n), (r+1))((r+1), (r))`

`vdots`

`text(When)\ \ n=n:\ \ ((n), (n))((n), (r))`
 

`text{Equating the coefficients:}`
 

`((n), (r))((r), (r)) + ((n), (r + 1))((r + 1), (r)) + … + ((n), (n))((n), (r)) =((n), (r)) 2^(n-r)`

`text(… as required)`

 

ii.   `text(Consider the possible combinations when,)`

`A\ text(selects 4:)\ \ ((23), (4))((4),(4))`

♦♦ Mean mark 32%.

`A\ text(selects 5:)\ \ ((23), (5))((5),(4))`

`A\ text(selects 6:)\ \ ((23), (6))((6),(4))`

`vdots`

`A\ text(selects 23:)\ \ ((23), (23))((23),(4))`
 

`:.\ text(Total possibilities)`

`=underbrace{((23), (4))((4),(4)) + ((23), (5))((5),(4)) + … + ((23), (23))((23),(4))}_text{Using part (i)}`

`=((23), (4)) 2^(23-4)`

`= ((23), (4)) 2^(19)`

Combinatorics, EXT1 2013 HSC 14b

  1. Write down the coefficient of `x^(2n)` in the binomial expansion of `(1 + x)^(4n)`.    (1 mark)

  2. Show that  
  3. `(1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k)) x^(2n-k)(x + 2)^(2n-k)`.   (2 marks)

  4. It is known that  
  5. `x^(2n-k) (x + 2)^(2n-k) = ((2n-k),(0)) 2^(2n-k) x^(2n-k) + ((2n-k),(1)) 2^(2n-k-1) x^(2n-k + 1)`
  6.     `+ ... + ((2n-k),(2n-k)) 2^0 x^(4n-2k)`.   (Do NOT prove this.)
  7. Show that
  8. `((4n),(2n)) = sum_(k = 0)^(n) 2^(2n-2k) ((2n),(k))((2n-k),(k))`.   (3 marks)
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i.    `((4n),(2n))`

ii.   `text(Proof)\ \ text{(See Worked Solutions)}`

iii.  `text(Proof)\ \ text{(See Worked Solutions)}`

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i.    `text(Find co-efficient of)\ x^(2n).`

`text(Expanding)\ (1+x)^(4n):`

`((4n),(0)) + ((4n),(1))x + ((4n),(2))x^2 + … + ((4n),(2n))x^(2n) + …`

`:.\ text(Co-efficient of)\ \ x^(2n)\ text(is)\ ((4n),(2n))`

 

ii.   `text(Show)\ (1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k)) x^(2n-k) (x + 2)^(2n-k)`

♦♦ Mean mark 30%.

`text(Using)\ (1 + x^2 + 2x)^(2n) = [x(x + 2) + 1]^(2n)`

`[x (x + 2) + 1]^(2n)`
`= ((2n),(0)) (x(x + 2))^(2n) + ((2n),(1)) (x(x + 2))^(2n-1) + … + ((2n),(2n))` 
`= ((2n),(0)) x^(2n)(x + 2)^(2n) + ((2n),(1)) x^(2n-1) (x + 2)^(2n-1) + … + ((2n),(2n))` 
`= sum_(k=0)^(2n) ((2n),(k)) x^(2n-k) (x + 2)^(2n-k)\ text(… as required.)`

 

iii.  `((4n),(2n))\ text(is the co-eff of)\ x^(2n)\ text(in expansion)\ (1+x)^(4x)`

`text(S)text(ince)\ (1 + x^2 + 2x)^(2n) = ((x+1)^2)^(2n) = (1 + x)^(4n)`

`=> ((4n),(2n))\ text(is co-efficient of)\ x^(2n)\ text(in expansion)\ (1 + x^2 + 2x)^(2n)`

 
`text(Using part)\ text{(ii):}`

`(1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k))\ x^(2n-k) (x + 2)^(2n-k)`
 

`text(Using the given identity,)\ x^(2n)\ text(co-efficients are:)`

♦♦♦ Toughest question in the 2013 exam with Mean mark 12%!
MARKER’S COMMENT: Stating `(1+x^2+2x)^(2n)=(1+x)^(4n)` received 1 full mark. Take note.
`k = 0,` `\ ((2n),(0))((2n-0),(0)) 2^(2n-0)`
`k = 1,` `\ ((2n),(1))((2n-1),(1)) 2^(2n-1-1)`
`vdots`  
`k = n,` `\ ((2n),(n))((2n-n),(n)) 2^(2n-n-n)`

 

`:.\ ((4n),(2n))`
 `= ((2n),(0))((2n),(0))2^(2n) + ((2n),(1))((2n-1),(1))2^(2n-2) + … + ((2n),(n))((n),(n)) 2^0`
` = sum_(k=0)^(n)\ ((2n),(k))((2n-k),(k)) 2^(2n-2k)\ \ \ \ text(… as required)`