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

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

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

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

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

♦ 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 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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\(p=2024, q=81 \)

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

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

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

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

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

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

 
\(\therefore p=2024, q=81 \)

Combinatorics, EXT1 2014 HSC 12d

Use the binomial theorem to show that

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

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

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`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 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)

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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`