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Combinatorics, EXT1 EQ-Bank 11

Find the term independent of \(x\) in the expansion of  \(\left(2 x^3+\dfrac{1}{x^4}\right)^7\).   (2 marks)

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\(560\)

Show Worked Solution

\(T_k=\ \text {General term of} \ \ \left(2 x^3+\dfrac{1}{x^4}\right)^7\)

\(T_k\) \(=\displaystyle \binom{7}{k}\left(2 x^3\right)^{7-k} \cdot\left(x^{-4}\right)^k\)
  \(=\displaystyle\binom{7}{k} \cdot 2^{7-k} \cdot x^{3(7-k)} \cdot x^{-4 k}\)
  \(=\displaystyle\binom{7}{k} \cdot 2^{7-k} \cdot x^{21-7 k}\)

 

\(\text{Independent term occurs when:}\)

\(x^{21-7 k}=x^0 \ \Rightarrow \ k=3\)

\(\therefore \text{Independent term}=\displaystyle \binom{7}{3} \cdot 2^4=560\)

Combinatorics, EXT1 A1 2025 HSC 13e

  1. The Pascal's triangle relation can be expressed as
    1. \(\displaystyle \binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}.\) (Do NOT prove this.)
  2. Show that \(\displaystyle \binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\).   (1 mark)

  3. Hence, or otherwise, prove that
  4. \(\displaystyle\binom{2000}{2000}+\binom{2001}{2000}+\binom{2002}{2000}+\cdots+\binom{2050}{2000}=\binom{2051}{2001}\).   (2 marks)
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i.    \(\displaystyle \binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}\ \ldots\ (1)\)

\(\text{Show:}\ \displaystyle \binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)

\(\text{Substitute} \ \ n=m+1 \ \ \text{and} \ \ r=R+1 \text{ into (1):}\)

\(\displaystyle\binom{m+1}{R+1}=\binom{m}{R}+\binom{m}{R+1}\)

\(\displaystyle\binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)
 

ii.    \(\text{Prove} \ \ \displaystyle \binom{2000}{2000}+\binom{2001}{2000}+\binom{2002}{2000}+\cdots+\binom{2050}{2000}=\binom{2051}{2001}\)

\(\text{Using part (i)}:\)

\(\operatorname{LHS}\) \(=1+\displaystyle \left[\binom{2002}{2001}-\binom{2001}{2001}\right]+\left[\binom{2003}{2001}-\binom{2002}{2001}\right]+\ldots\)
  \(\quad \quad \quad +\displaystyle \left[\binom{2050}{2001}-\binom{2049}{2001}\right]+\left[\binom{2051}{2001}-\binom{2050}{2001}\right]\)
  \(=1-1+\displaystyle \binom{2051}{2001}\)
  \(=\displaystyle \binom{2051}{2001}\)
Show Worked Solution

i.    \(\displaystyle \binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}\ \ldots\ (1)\)

\(\text{Show:}\ \displaystyle \binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)

\(\text{Substitute} \ \ n=m+1 \ \ \text{and} \ \ r=R+1 \text{ into (1):}\)

\(\displaystyle\binom{m+1}{R+1}=\binom{m}{R}+\binom{m}{R+1}\)

\(\displaystyle\binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)

♦ Mean mark (a) 43%.

ii.    \(\text{Prove} \ \ \displaystyle \binom{2000}{2000}+\binom{2001}{2000}+\binom{2002}{2000}+\cdots+\binom{2050}{2000}=\binom{2051}{2001}\)

\(\text{Using part (i)}:\)

\(\operatorname{LHS}\) \(=1+\displaystyle \left[\binom{2002}{2001}-\binom{2001}{2001}\right]+\left[\binom{2003}{2001}-\binom{2002}{2001}\right]+\ldots\)
  \(\quad \quad \quad +\displaystyle \left[\binom{2050}{2001}-\binom{2049}{2001}\right]+\left[\binom{2051}{2001}-\binom{2050}{2001}\right]\)
  \(=1-1+\displaystyle \binom{2051}{2001}\)
  \(=\displaystyle \binom{2051}{2001}\)
♦♦ Mean mark (b) 37%.

Combinatorics, EXT1 A1 EQ-Bank 15

\(\left(\sqrt{5}-2 \right)^5=x+y \sqrt{5}\)

Find the values of \(x\) and \(y\) using binomial expansion.   (2 marks)

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\(x=-682,  y=305\)

Show Worked Solution

  \( \left(\sqrt{5}-2\right)^5\) \(={ }^5 C_0 \cdot\left(\sqrt{5}\right)^5+{ }^5 C_1\left(\sqrt{5}\right)^4 \cdot(-2)+{ }^5 C_2\left(\sqrt{5}\right)^3 \cdot(-2)^2\)
    \(+{ }^5 C_3\left(\sqrt{5}\right)^2 \cdot(-2)^3 +{ }^5 C_4\left(\sqrt{5}\right) \cdot(-2)^4+{ }^5 C_5(-2)^5 \)
    \( = 25 \sqrt{5}-250+200 \sqrt{5}-400+80 \sqrt{5}-32\)
    \(=  305 \sqrt{5}-682 \)

 
\(\therefore x=-682,  y=305\)

Combinatorics, EXT1 A1 2023 HSC 12d

It is known that  \({ }^n C_r={ }^{n-1} C_{r-1}+{ }^{n-1} C_r\)  for all integers such that  \(1 \leq r \leq n-1\). (Do NOT prove this.)

Find ONE possible set of values for \(p\) and \(q\) such that

\({ }^{2022} C_{80}+{ }^{2022} C_{81}+{ }^{2023} C_{1943}={ }^p C_q\)  (2 marks)

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\(p=2024, q=81 \)

Show Worked Solution

\({ }^{2022} C_{80}+{ }^{2022} C_{81}+{ }^{2023} C_{1943}={ }^p C_q\)

\(\text{Using the known relationship:} \)

\({ }^{2022} C_{80}+{ }^{2022} C_{81} = { }^{2023} C_{81}\ \ …\ (1)\)

\(\text{Also, since}\ \ { }^n C_r={ }^n C_{n-r} \)

\({ }^{2023} C_{1943} = { }^{2023} C_{2023-1943} = { }^{2023} C_{80}\ \ …\ (2)\) 

\({ }^{2022} C_{80}+{ }^{2022} C_{81}+{ }^{2023} C_{1943}\) \(={ }^{2023} C_{81}+{ }^{2023} C_{1943}\ \ \text{(see (1) above)}\)  
  \(={ }^{2023} C_{81}+{ }^{2023} C_{80}\ \ \text{(see (2) above)} \)  
  \(={ }^{2024} C_{81} \)  

 
\(\therefore p=2024, q=81 \)

Combinatorics, EXT1 A1 2022 HSC 11c

Find the coefficients of `x^(2)` and `x^(3)` in the expansion of `(1-(x)/(2))^(8)`.  (2 marks)

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`x^2: 7, \ x^3:-7`

Show Worked Solution

`text{General Term:}\ (1-(x)/(2))^(8) `

`T_k` `=((8),(k))(1)^(8-k)(- x/2)^k`  
  `=((8),(k))(- 1/2)^k x^k`  

 
`text{Coefficient of}\ \ x^2 = ((8),(2))(- 1/2)^2=7`

`text{Coefficient of}\ \ x^3 = ((8),(3))(- 1/2)^3=-7`

Combinatorics, EXT1 A1 2021 HSC 11b

Expand and simplify  `(2a - b)^4`.  (2 marks)

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`16a^4 – 32a^3b + 24a^2b^2 – 8ab^3 + b^4`

Show Worked Solution

`(2a – b)^4`

`= (2a)^4 + \ ^4C_1(2a)^3(-b) + \ ^4C_2(2a)^2(-b)^2 + \ ^4C_3(2a)(-b)^3 + (-b)^4`

`= 16a^4 – 4 · 8a^3b + 6 · 4a^2b^2 – 4 · 2ab^3 + b^4`

`= 16a^4 – 32a^3b + 24a^2b^2 – 8ab^3 + b^4`

Combinatorics, EXT1 A1 2020 HSC 14a

  1. Use the identity `(1 + x)^(2n) = (1 + x)^n(1 + x)^n`

     

    to show that
     
        `((2n),(n)) = ((n),(0))^2 + ((n),(1))^2 + … + ((n),(n))^2`,
     
    where `n` is a positive integer.  (2 marks)

  2. A club has `2n` members, with `n` women and `n` men.

     

    A group consisting of an even number `(0, 2, 4, …, 2n)` of members is chosen, with the number of men equal to the number of women.
     
    Show, giving reasons, that the number of ways to do this is `((2n),(n))`.  (2 marks)

  3. From the group chosen in part (ii), one of the men and one of the women are selected as leaders.

     

    Show, giving reasons, that the number of ways to choose the even number of people and then the leaders is
     

     

        `1^2 ((n),(1))^2 + 2^2((n),(2))^2 + … + n^2((n),(n))^2`.  (2 marks)

  4. The process is now reversed so that the leaders, one man and one woman, are chosen first. The rest of the group is then selected, still made up of an equal number of women and men.

     

    By considering this reversed process and using part (ii), find a simple expression for the sum in part (iii).  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `n^2 xx \ ^(2n-2)C_(n-1)`
Show Worked Solution

i.   `text(Expand)\ \ (1 + x)^(2n):`

♦♦ Mean mark part (i) 26%.

`\ ^(2n)C_0 + \ ^(2n)C_1 x^2 + … + \ ^(2n)C_n x^n + … \ ^(2n)C_(2n) x^(2n)`

`=> text(Coefficient of)\ \ x^n = \ ^(2n)C_n`
 

`text(Expand)\ \ (1 + x)^n (1 + x)^n:`

`[\ ^nC_0 + \ ^nC_1 x + … + \ ^nC_n x^n][\ ^nC_0 + \ ^nC_1 x + … + \ ^nC_n x^n]`

`=> \ text(Coefficient of)\ \ x^n`

`= \ ^nC_0 · \ ^nC_n + \ ^nC_1 · \ ^nC_(n-1) + … + \ ^nC_(n-1) · \ ^nC_1 + \ ^nC_n · \ ^nC_0`

`= (\ ^nC_0)^2 + (\ ^nC_1)^2 + … + (\ ^nC_(n-1))^2 + (\ ^nC_n)^2\ \ \ (\ ^nC_k = \ ^nC_(n-k))`
 

`text(Equating coefficients:)`

`\ ^(2n)C_n = (\ ^nC_0)^2 + (\ ^nC_1)^2 + … + (\ ^nC_n)^2`

♦♦ Mean mark part (ii) 23%.

 

ii.   `text(Number of men = Number of women)\ \ (M = W)`

`text(If)\ \ M = W = 0:`  `text(Ways) = \ ^nC_0 · \ ^nC_0 = (\ ^nC_0)^2`
`text(If)\ \ M = W = 1:`  `text(Ways) = \ ^nC_1 · \ ^nC_1 = (\ ^nC_1)^2`

`vdots`

`text(If)\ \ M = W = n:  text(Ways) = \ ^nC_n · \ ^nC_n = (\ ^nC_n)^2`
 

`:.\ text(Total combinations)`

`= (\ ^nC_0)^2 + (\ ^nC_1)^2 + … + (\ ^nC_n)^2`

`= \ ^(2n)C_n\ \ \ text{(from part (i))}`

 

iii.   `text(Let)\ \ M_L = text(possible male leaders)`

♦♦ Mean mark part (iii) 26%.

`text(Let)\ \ W_L = text(possible female leaders)`

`text(If)\ \ M = W = 0 => text(no leaders)`

`text(If)\ \ M = W = 1:  text(Ways) = \ ^nC_1 xx M_L xx \ ^nC_1 xx W_L = 1^2 (\ ^nC_1)^2`

`text(If)\ \ M = W = 2:  text(Ways) = \ ^nC_2 xx 2 xx \ ^nC_2 xx 2 = 2^2 (\ ^nC_2)^2`

`vdots`

`text(If)\ \ M = W = n:  text(Ways) = \ ^nC_n xx n xx \ ^nC_2 xx n = n^2 (\ ^nC_n)^2`
 

`:.\ text(Total combinations)`

`= 1^2(\ ^nC_1)^2 + 2^2(\ ^nC_2)^2 + … + n^2(\ ^nC_n)^2`

♦♦♦ Mean mark part (iv) 16%.

 

iv.  `text(If)\ \ M = W = 1:  text(Ways)` `= M_L xx \ ^(n-1)C_0 xx W_L xx \ ^(n-1)C_0`
    `= n xx \ ^(n-1)C_0 xx n xx \ ^(n-1)C_0`
    `= n^2(\ ^(n-1)C_0)^2`
`text(If)\ \ M = W = 2:  text(Ways)` `= n xx \ ^(n-1)C_1 xx n xx \ ^(n-1)C_1`
  `= n^2(\ ^(n-1)C_1)^2`

`vdots`

`text(If)\ \ M = W = n:\ text(Ways)` `= n xx \ ^(n-1)C_(n-1) xx n xx \ ^(n-1)C_(n-1)`
  `= n^2(\ ^(n-1)C_(n-1))^2`

 
`:.\ text(Total combinations)`

`= n^2(\ ^(n-1)C_0)^2 + n^2(\ ^(n-1)C_1)^2 + … + n^2(\ ^(n-1)C_(n-1))^2`

`= n^2 xx \ ^(2(n-1))C_(n-1)\ \ \ text{(using part (i))}`

`= n^2 xx \ ^(2n-2)C_(n-1)`

Combinatorics, EXT1 A1 2019 MET1 8

A fair standard die is rolled 50 times. Let `W` be a random variable with binomial distribution that represents the number of times the face with a six on it appears uppermost.

  1. Write down the expression for  `P(W = k)`, where  `k in {0, 1, 2, …, 50}`.   (1 mark)

  2. Show that  `(P(W = k + 1))/(P(W = k)) = (50-k)/(5(k + 1))`.   (2 marks)

Show Answers Only
  1. `\ ^50 C_k ⋅ (1/6)^k ⋅ (5/6)^(50-k)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

a.  `P(W = k) = \ ^50 C_k ⋅ (1/6)^k ⋅ (5/6)^(50-k)`

 

b.   `(P(W = k + 1))/(P(W = k))` `= (\ ^50C_(k+1) ⋅ (1/6)^(k+1) ⋅ (5/6)^(49-k))/(\ ^ 50C_k ⋅ (1/6)^k ⋅ (5/6)^(50-k))`
    `= ((50!)/((49-k)!(k + 1)!) ⋅ (1/6))/((50!)/((50-k)! k!) ⋅ (5/6))`
    `= ((50-k)!k!)/(5(49-k)!(k + 1)!)`
    `= (50-k)/(5(k + 1))`

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)`

Show Worked Solution

`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 9

`(2 - sqrt3)^5 = a + bsqrt3`.

Evaluate `a` and `b` using a binomial expansion.  (2 marks)

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`a = 362, \ b = -209`

Show Worked Solution
`(2 – sqrt3)^5` `= \ ^5C_0 *2^5 + \ ^5C_1* 2^4(−sqrt3) + \ ^5C_2 *2^3(−sqrt3)^2 + \ ^5C_3 *2^2(−sqrt3)^3`
               `+ \ ^5C_4 *2(−sqrt3)^4 + \ ^5C_5(−sqrt3)^5`
  `= 32 – 80sqrt3 + 240 – 120sqrt3 + 90 – 9sqrt3`
  `= 362 – 209sqrt3`

 
`:. a = 362, \ b = -209`

Combinatorics, EXT1 A1 EQ-Bank 7

Show `\ ^nC_k = \ ^(n-1)C_(k-1) + \ ^(n-1)C_k`.   (2 marks)

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`text(See Worked Solutions)`

Show Worked Solution

`text(LHS) = (n!)/((n-k)!k!)`

`text(RHS)` `= ((n-1)!)/((n-1-(k-1))!(k-1)!) + ((n-1)!)/((n-1-k)!k!)`
  `= ((n-1)!k)/((n-k)!(k-1)!k) + ((n-1)!(n-k))/((n-k-1)!(n-k)k!)`
  `= ((n-1)!k)/((n-k)!k!) + ((n-1)!(n-k))/((n-k)!k!)`
  `= ((n-1)!(k + n-k))/((n-k)!k!)`
  `= (n!)/((n-k)!k!)`
  `=\ text(LHS)`

Combinatorics, EXT1 A1 2019 13b

In the expansion of  `(5x + 2)^20`, the coefficients of  `x^r`  and  `x^(r + 1)`  are equal.

What is the value of  `r` ?  (3 marks)

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`14`

Show Worked Solution

`(5x + 2)^20`

COMMENT: Arithmetic becomes easier by expanding  `(2 + 5x)^20`.

`text(General term:)\ (2 + 5x)^20`

`T_r` `= \ ^20C_r · 2^(20 – r) · (5x)^r`
  `= \ ^20C_r · 2^(20 – r) · 5^r · x^r`
`T_(r + 1)` `=\ ^20C_(r + 1) · 2^(19 – r) · 5^(r + 1) · x^(r + 1)`

 

`text(Equating co-efficients:)`

`(20!)/(r!(20 – r)!) · 2^(20 – r) · 5^r` `= (20!)/((r + 1)!(19 – r)!) · 2^(19 – r) · 5 ^(r + 1)`
`2/(20 – r)` `= 5/(r + 1)`
`2r + 2` `= 100 – 5r`
`7r` `= 98`
`r` `= 14`

Combinatorics, EXT1 A1 EQ-Bank 3

Find the coefficient of  `x^4`  in the expansion of  `(x^2 - 3/x)^5`.  (2 marks)

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`90`

Show Worked Solution

`text(General term:)`

`T_k` `= \ ^5C_k(x^2)^(5 – k) · (−3/x)^k`
  `= \ ^5C_k · x^(10 – 2k)(−3)^k · x^(−k)`
  `= \ ^5C_k · x^(10 – 3k) · (−3)^k`

 
`text(Coefficient of)\ \ x^4\ \ text(occurs when)`

`10 – 3k` `= 4`
`3k` `= 6`
`k` `= 2`

 
`:.\ text(Coefficient of)\ \ x^4`

`= \ ^5C_2·(−3)^2`

`= 90`

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)`

Show Worked Solution

`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 A1 SM-Bank 1

`(1-2sqrt2)^6 = x + ysqrt2`

Evaluate the value of `x` and `y`.  (2 marks)

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`x = 1593, \ y =-1100`

Show Worked Solution

`text(Using the binomial expansion:)`

`(1-2sqrt2)^6`

`= \ ^6C_0 + \ ^6C_1(-2sqrt2) + \ ^6C_2(-2sqrt2)^2 + \ ^6C_3(-2sqrt2)^3 + \ ^6C_4(-2sqrt2)^4`

`+ \ ^6C_5(-2sqrt2)^5 + \ ^6C_6(-2sqrt2)^6`
 

`= 1-12sqrt2 + 120 -320sqrt2 + 960-768sqrt2 + 512`

`= 1593-1100sqrt2`
 

`:.x = 1593, \ y =-1100`

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`

Show Worked Solution

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 A1 2017 HSC 9 MC

When expanded, which expression has a non-zero constant term?

A.     `(x + 1/(x^2))^7`

B.     `(x^2 + 1/(x^3))^7`

C.     `(x^3 + 1/(x^4))^7`

D.     `(x^4 + 1/(x^5))^7`

Show Answers Only

`C`

Show Worked Solution

`text(Consider the general term for option)\ A:`

`T_k` `= \ ^7C_k · x^(7 – k) · x^(−2k)`
  `= \ ^7C_k · x^(7 – 3k)`

 
`text(Non zero constant term occurs when)`

`7 – 3k` `= 0`
`k` `= 7/3 => text(no terms exists)\ (k\ text{not integer)}`

 
`text(Consider option)\ C:`

`T_k` `= \ ^7C_k · x^(3(7 – k)) · x^(−4k)`
  `= \ ^7C_k · x^(21 – 7k)`
`21 – 7k` `= 0`
`k` `= 3`

 
`:.\ text(Non-zero constant term exists)`

`text(since)\ k\ text(is an integer)`

`⇒C`

Combinatorics, EXT1′ A1 2016 HSC 6 MC

Let  `p(x) = 1 + x + x^2 + x^3 + … + x^12.`

What is the coefficient of `x^8` in the expansion of  `p (x + 1)?`

  1. `1`
  2. `495`
  3. `715`
  4. `1287`
Show Answers Only

`=> C`

Show Worked Solution

`p(x + 1) = 1 + (x + 1) + (x + 1)^2 + … + (x + 1)^12`

`text(Coefficient of)\ x^8`

`=\ ^12C_8 +\ ^11C_8 +\ ^10C_8 +\ ^9C_8 +\ ^8C_8`

`= 715`

`=> C`

Combinatorics, EXT1 2008 HSC 6c

Let `p` and `q` be positive integers with  `p ≤ q`.

  1. Use the binomial theorem to expand `(1 + x) ^(p+ q)`, and hence write down the term of
  2. `((1 + x)^(p + q))/(x^q)`  which is independent of  `x`.   (2 marks)

  3. Given that  `((1 + x)^(p + q))/(x^q) = (1 + x)^p(1 + 1/x)^q`,
  4. apply the binomial theorem and the result of part (i) to find a simpler expression for
  5. `1 + ((p),(1))((q),(1)) + ((p),(2))((q),(2)) + … + ((p),(p))((q),(p))`.   (3 marks)

Show Answers Only

i.    `\ ^(p + q)C_q`

ii.   `\ ^(p + q)C_q`

Show Worked Solution

i.    `(1 + x)^(p + q)`

`=\ ^(p + q)C_0 +\ ^(p + q)C_1 x + … +\ ^(p + q)C_q x^q + … +\ ^(p + q)C_(p + q) · x^(p + q)`

`:.\ text(Independent term of)\ ((1 + x)^(p + q))/(x^q)`

`= (\ ^(p + q)C_q·x^q)/(x^q)`

`=\ ^(p + q)C_q`

 

ii.   `(1 + x)^p(1 + 1/x)^q`

`= (\ ^pC_0 +\ ^pC_1 x + … +\ ^pC_p x^p)`

`xx (\ ^qC_0 +\ ^qC_1 · 1/x + … +\ ^qC_p · 1/(x^p) + … +\ ^qC_q · 1/(x^q))`
 

`text(The independent term in this expansion)`

`=\ ^pC_0 ·\ ^qC_0 +\ ^pC_1 ·\ ^qC_1 + … +\ ^pC_p ·\ ^qC_p\ text{(since}\ p ≤ q)`

`= 1 +\ ^pC_1 ·\ ^qC_1 + … +\ ^pC_p ·\ ^qC_p`
 

`text(S)text(ince)\ ((1 + x)^(p + q))/(x^q) = (1 + x)^p(1 + 1/x)^q,\ text(the independent)`

`text(terms are equal.)`
 

`:.\ ^(p + q)C_q\ text(is a simpler expression for)`

`1 +\ ^pC_1 ·\ ^qC_1 +\ ^pC_2 ·\ ^qC_2 + … +\ ^pC_p ·\ ^qC_p`

Combinatorics, EXT1 A1 2005 HSC 2b

Use the binomial theorem to find the term independent of  `x`  in the expansion of

`(2x-1/x^2)^12.`   (3 marks)

Show Answers Only

`((12), (4)) * 2^8 * (-1)^4 = 126\ 720`

Show Worked Solution

`T_k =\ text(General term of)\ \ (2x-1/x^2)^12`

`T_k` `= ((12), (k)) (2x)^(12-k) * (-1)^k * (x^-2)^k`
  `= ((12), (k)) * 2^(12-k) * x^(12-k) * (-1)^k * x^(-2k)`
  `= ((12), (k)) * 2^(12-k) * (-1)^k * x^(12-3k)`

 
`\text{Independent term occurs when:}`

`x^(12-3k)= x^0\ \ =>\ \ k=4`
 

`:.\ text(Independent term is)`

`((12), (4)) * 2^8 * (-1)^4 = 126\ 720`

Combinatorics, EXT1 A1 2015 HSC 14c

Two players `A` and `B` play a series of games against each other to get a prize. In any game, either of the players is equally likely to win.

To begin with, the first player who wins a total of 5 games gets the prize.

  1. Explain why the probability of player `A` getting the prize in exactly 7 games is  `((6),(4))(1/2)^7`.   (1 mark)

  2. Write an expression for the probability of player `A` getting the prize in at most 7 games.   (1 mark)

  3. Suppose now that the prize is given to the first player to win a total of `(n + 1)` games, where `n` is a positive integer.

     

    By considering the probability that `A` gets the prize, prove that
     

  4. `((n),(n))2^n + ((n + 1),(n))2^(n − 1) + ((n + 2),(n))2^(n − 2) + … + ((2n),(n)) = 2^(2n)`.   (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `\ ^4C_4(1/2)^5 +\ ^5C_4(1/2)^6 +\ ^6C_4(1/2)^7`
  3. `text{Proof (See Worked Solutions.)}`
Show Worked Solution

i.   `text(To win exactly 7 games, player)\ A`

♦♦♦ Mean mark 19%.

`text(must win the 7th game.)`
 

`:.P(A\ text{wins in 7 games)}`

`=\ ^6C_4 · (1/2)^4(1/2)^2 xx 1/2`

`=\ ^6C_4(1/2)^7`

 

ii.  `Ptext{(wins in at most 7 games)}`

♦♦♦ Mean mark 23%.

`=Ptext{(wins in 5, 6 or 7 games)}`

`=\ ^4C_4(1/2)^4 xx 1/2 +\ ^5C_4(1/2)^4(1/2) xx 1/2 +\ ^6C_4(1/2)^7`

`=\ ^4C_4(1/2)^5 +\ ^5C_4(1/2)^6 +\ ^6C_4(1/2)^7`

♦♦♦ Mean mark part (iii) 9%.

 

iii. `text(Prove that)`

`\ ^nC_n · 2^n +\ ^(n + 1)C_n · 2^(n-1) + … +\ ^(2n)C_n = 2^(2n)`

`P(A\ text(wins in)\ (n + 1)\ text{games)}`

`=\ ^nC_n(1/2)^(n + 1) +\ ^(n + 1)C_n(1/2)^(n + 2) + … +\ ^(2n)C_n(1/2)^(2n + 1)`
 

`text{One player must have won after (2n + 1) games are played.}`

`text(S)text(ince each player has an equal chance,)`
 

`\ ^nC_n(1/2)^(n + 1) +\ ^(n + 1)C_n(1/2)^(n + 2) + … +\ ^(2n)C_n(1/2)^(2n + 1) = 1/2`
 

`text(Multiply both sides by)\ 2^(2n + 1):`
 

`\ ^nC_n2^(-(n + 1)) · 2^(2n + 1) +\ ^(n + 1)C_n · 2^(-(n + 2)) · 2^(2n + 1) + …`

`… +\ ^(2n)C_n · 2^(-(2n + 1)) · 2^(2n + 1) = 2^(-1) · 2^(2n + 1)`
 

`:. \ ^nC_n · 2^n +\ ^(n + 1)C_n · 2^(n-1) + … +\ ^(2n)C_n = 2^(2n)`

Combinatorics, EXT1 A1 2015 HSC 13b

Consider the binomial expansion
 

`(2x + 1/(3x))^18 = a_0x^(18) + a_1x^(16) + a_2x^(14) + …`
 

where `a_0, a_1, a_2`, . . . are constants.

  1. Find an expression for `a_2`.  (2 marks)

  2. Find an expression for the term independent of `x`.  (2 marks)

Show Answers Only
  1. `(\ ^18C_2 · 2^(16))/(3^2)`
  2. `(\ ^(18)C_9 · 2^9)/(3^9)`
Show Worked Solution

i.   `text(Need co-efficient of)\ x^(14)`

`text(General term of)\ (2x + 1/(3x))^(18)`

`T_k` `= \ ^(18)C_k(2x)^(18 − k) · (1/(3x))^k`
  `= \ ^(18)C_k · 2^(18 − k) · x^(18 − k) · 3^(−k) · x^(−k)`
  `= \ ^(18)C_k · 2^(18 − k) · 3^(−k) · x^(18 − 2k)`

 

`a_2\ text(occurs when:)`

`18 − 2k` `= 14`
`2k` `= 4`
`k` `= 2`

 

`:.a_2` `= \ ^(18)C_2 · 2^(18 − 2) · 3^(−2)`
  `= (\ ^(18)C_2 · 2^(16))/(3^2)`

 

ii.  `text(Independent term occurs when:)`

`18 − 2k` `= 0`
`2k` `= 18`
`k` `= 9`

 
`:.\ text(Independent term)`

`= \ ^(18)C_9 · 2^(18− 9) · 3^(−9)`

`= (\ ^(18)C_9 · 2^9)/(3^9)`

Combinatorics, EXT1 A1 2008 HSC 1d

Find an expression for the coefficient of  `x^8 y^4`  in the expansion of  `(2x + 3y)^12`.   (2 marks) 

Show Answers Only

`10\ 264\ 320`

Show Worked Solution

`text(Find co-efficient of)\ x^8 y^4 :`

MARKER’S COMMENT: More errors were made by students who used `T_(k+1)` as the general term rather than `T_k` (both are possible). The Worked Solution uses the more successful approach.

`T_k =\ text(General term of)\ (2x + 3y)^12 `

`T_k` `= ((12),(k)) (2x)^(12 – k) * (3y)^k`
  `= ((12),(k)) * 2^(12 – k) * 3^k * x^(12 – k) * y^k`

 

`x^8 y^4\ text(occurs when)\ k = 4`

`T_4` `= ((12),(4)) * 2^(12 – 4) * 3^4 * x^8 y^4`
   
`:.\ text(Co-efficient of)\ x^8y^4`
  `= ((12),(4)) * 2^8 * 3^4`
  `= 10\ 264\ 320`

Combinatorics, EXT1 2014 HSC 12d

Use the binomial theorem to show that

`0 = ((n),(0))-((n),(1)) + ((n),(2))-... + (-1)^n ((n),(n))`.   (2 marks) 

Show Answers Only

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

Show Worked Solution

`text(Prove)`

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

`text(Using binomial expansion:)`

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

`text(Let)\ \ x = -1`

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

Combinatorics, EXT1 A1 2014 HSC 3 MC

What is the constant term in the binomial expansion of   `(2x - 5/(x^3))^12`?

  1. `((12),(3)) 2^9 5^3`
  2. `((12),(9)) 2^3 5^9`
  3. `-((12),(3)) 2^9 5^3`
  4. `-((12),(9)) 2^3 5^9`
Show Answers Only

`C`

Show Worked Solution

`text(General term)`

`T_k` `= ((12),(k)) (2x)^(12-k) * (-1)^k *(5x^(-3))^k`
  `= ((12),(k)) 2^(12-k) * x^(12-k) * (-1)^k * 5^k * x^(-3k)`
  `= ((12),(k)) (-1)^k * 2^(12-k) * 5^k * x^(12-4k)`

 
`text(Constant term when)`

`12 – 4k` `= 0`
`k` `= 3`

 
`:.\ text(Constant term)`

`=((12),(3)) (-1)^3 * 2^9 * 5^3`

`= – ((12),(3)) * 2^9 * 5^3`
 

`=>  C`

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)
Show Answers Only

i.    `((4n),(2n))`

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

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

Show Worked Solution

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)`

Combinatorics, EXT1 2010 HSC 7b

The binomial theorem states that

`(1 + x)^n = ((n),(0)) + ((n),(1))x + ((n),(2))x^2 + ((n),(3))x^3 + ... + ((n),(n))x^n.` 

  1. Show that  `2^n = sum_(k = 0)^n ((n),(k))`.   (1 mark)

  2. Hence, or otherwise, find the value of
  3. `((100),(0)) + ((100),(1)) + ((100),(2)) + ... + ((100),(100))`.   (1 mark)

Show Answers Only

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

ii.   `2^100`

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

Show Worked Solution

i.   `(1 + x)^n = ((n),(0)) + ((n),(1))x + … + ((n),(n))x^n`

`text(Let)\ \ x = 1:`

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

 

ii.   `text{Using part (i), let}\ \ n=100:`

`((100),(0)) + ((100),(1)) + ((100),(2)) + … + ((100),(100)) = 2^100`

Combinatorics, EXT1 A1 2011 HSC 2c

Find an expression for the coefficient of  `x^2`  in the expansion of  `(3x - 4/x)^8`.     (2 marks)

Show Answers Only

 `-870\ 912`

Show Worked Solution

`(3x – 4/x)^8 = sum_(k=0)^8\ ^8C_k *(3x)^(8\ – k) *(–4/x)^k`

`text(General term)` `=\ ^8C_k * 3^(8\ – k) * x^(8\ – k) * (–1)^k*4^k * x^(-k)`
  `=\ ^8C_k*(–1)^k * 3^(8\ – k) * 4^k * x^(8\ – 2k)`

 
`text(Co-efficient of)\ \ x^2=2\ \ text(occurs when,)`

` 8 – 2k` `= 2`
`2k` `= 6`
`k` `= 3`

 
`:.\ text(Co-efficient of)\ \ x^2`

`=\ ^8C_3 * (–1)^3*3^5 * 4^3`
`= -56 xx 243 xx 64`
`= -870\ 912`

Combinatorics, EXT1 A1 2012 HSC 11f

 

  1. Use the binomial theorem to find an expression for the constant term in the expansion of 
     
         `(2x^3 - 1/x)^12`.   (2 marks)

  2. For what values of  `n`  does  `(2x^3 - 1/x)^n`  have a non-zero constant term?    (1 mark)

Show Answers Only
  1. `-1760`
  2. `n\ text(must be a multiple of 4)`
Show Worked Solution

i.  `text(General term)`

`\ ^12C_k * (2x^3)^(12\ – k) (-1/x)^k` 

`=\ ^12C_k * (–1)^k *2^(12\ – k) * x^(36\ – 3k)  * x^(-k)`

`=\ ^12C_k * (–1)^k*2^(12\ – k) * x^(36\ – 4k)`
 

`text(Constant term occurs when)`

`36\ – 4k` `= 0`
`k` `= 9`

 

`:.\ text(Constant term)` `=\ ^12C_9 * (–1)^9*2^3`
  `= – (12!)/(3!9!) xx 8`
  `= – 1760`

 

ii.  `text(General term of)\ (2x^3\ – 1/x)^n`

♦♦♦ Mean mark 16%.

`\ ^nC_k * (2x^3)^(n\ – k) (–1/x)^k`

`=\ ^nC_k * 2^(n\ – k) * x^(3n\ – 3k) * (–1)^k * x^(-k)`

`=\ ^nC_k * (–1)^k*2^(n\ -k) * x^(3n\ – 4k)`

 

`text(Constant term when)\ \ 3n\ – 4k = 0.`

`text(i.e.)\ \ k=3/4n`
 

`text(S)text(ince)\ n\ text(and)\ k\ text(must be integers,)\ \ n\ \ text(must)`

`text(be a multiple of 4.)`