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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 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)

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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 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)

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  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)`