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

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

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

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

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

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

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

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

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 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)

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  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 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 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)`