From b51b0e7b005b451bb0f9f90c4a2ea9622ee6becf Mon Sep 17 00:00:00 2001 From: John Lu Date: Thu, 18 Jul 2024 00:05:12 -0400 Subject: [PATCH] added more solutions to 2014 fall final --- .../comp2804/2014-fall-final/10/solution.md | 38 +++++----- .../comp2804/2014-fall-final/12/solution.md | 31 ++++---- .../comp2804/2014-fall-final/13/solution.md | 8 ++ .../comp2804/2014-fall-final/14/solution.md | 16 ++-- .../comp2804/2014-fall-final/15/solution.md | 20 ++--- .../comp2804/2014-fall-final/16/solution.md | 20 ++--- .../comp2804/2014-fall-final/17/solution.md | 32 ++++---- .../comp2804/2014-fall-final/18/solution.md | 32 ++++---- .../comp2804/2014-fall-final/19/solution.md | 28 ++++--- .../comp2804/2014-fall-final/2/solution.md | 30 ++++---- .../comp2804/2014-fall-final/20/solution.md | 3 + .../comp2804/2014-fall-final/21/solution.md | 73 +++++++++---------- .../comp2804/2014-fall-final/22/solution.md | 47 ++++++------ .../comp2804/2014-fall-final/23/solution.md | 5 ++ .../comp2804/2014-fall-final/24/solution.md | 24 +++--- .../comp2804/2014-fall-final/3/solution.md | 9 +++ .../comp2804/2014-fall-final/4/solution.md | 37 ++++++---- .../comp2804/2014-fall-final/5/solution.md | 2 +- .../comp2804/2014-fall-final/6/solution.md | 9 +++ .../comp2804/2014-fall-final/7/solution.md | 8 ++ .../comp2804/2014-fall-final/8/solution.md | 18 +++-- .../comp2804/2014-fall-final/9/solution.md | 8 +- 22 files changed, 286 insertions(+), 212 deletions(-) diff --git a/src/content/questions/comp2804/2014-fall-final/10/solution.md b/src/content/questions/comp2804/2014-fall-final/10/solution.md index 5eaa4aed..1cc8fd33 100644 --- a/src/content/questions/comp2804/2014-fall-final/10/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/10/solution.md @@ -1,24 +1,28 @@ Let's calculate $f(1)$ first -$f(1) = f(0) + 8 \cdot 1 - 2 $ + +$f(1) = f(0) + 8 \cdot 1 - 2 $ + $f(1) = -17 + 8 - 2 $ + $f(1) = -11 $ + The correct answer is $ f(n) = 4n^{2} + 2n - 17 $ because it is the only one that gives the correct value for $ f(1) $ diff --git a/src/content/questions/comp2804/2014-fall-final/12/solution.md b/src/content/questions/comp2804/2014-fall-final/12/solution.md index 646cf304..55ffffc5 100644 --- a/src/content/questions/comp2804/2014-fall-final/12/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/12/solution.md @@ -1,17 +1,20 @@ -$ |A \cup B \cup C| = |A| + |B| + |C| $ -$ |A \cup B \cup C| = \binom{5}{2} + \binom{4}{2} + \binom{6}{2} $ + +$ |A \cup B \cup C| = |A| + |B| + |C| $ + +$ |A \cup B \cup C| = \binom{5}{2} + \binom{4}{2} + \binom{6}{2} $ + $ Pr(A \cup B \cup C) = \frac{\binom{5}{2} + \binom{4}{2} + \binom{6}{2}}{\binom{15}{2}} $ diff --git a/src/content/questions/comp2804/2014-fall-final/13/solution.md b/src/content/questions/comp2804/2014-fall-final/13/solution.md index fe0fa201..4497bdc8 100644 --- a/src/content/questions/comp2804/2014-fall-final/13/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/13/solution.md @@ -1,9 +1,17 @@ Let A = Event that Annie, Boris, and Charlie have the same birthday + Annie has a birthday on any of the 365 days: 365 + Boris would only then have a birthday on the same day as Annie: 1 + Charlie would only then have a birthday on the same day as Annie and Boris: 1 + $ |A| = 365 \cdot 1 \cdot 1 $ + Let S = All possible outcomes + $ |S| = 365^3 $ + $ Pr(A) = \frac{365}{365^3} $ + $ Pr(A) = \frac{1}{365^2} $ diff --git a/src/content/questions/comp2804/2014-fall-final/14/solution.md b/src/content/questions/comp2804/2014-fall-final/14/solution.md index c283f6c2..709b6797 100644 --- a/src/content/questions/comp2804/2014-fall-final/14/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/14/solution.md @@ -1,12 +1,12 @@ Let's go through each option and explain diff --git a/src/content/questions/comp2804/2014-fall-final/15/solution.md b/src/content/questions/comp2804/2014-fall-final/15/solution.md index 74ab9c59..0e187dc6 100644 --- a/src/content/questions/comp2804/2014-fall-final/15/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/15/solution.md @@ -1,12 +1,14 @@ -$ Pr(A|B) = \frac{|A \cap B|}{|B|} $ + +$ Pr(A|B) = \frac{|A \cap B|}{|B|} $ + $ Pr(A|B) = \frac{\binom{4}{2}}{\binom{5}{2} + \binom{4}{2}} $ diff --git a/src/content/questions/comp2804/2014-fall-final/16/solution.md b/src/content/questions/comp2804/2014-fall-final/16/solution.md index 9d550bc4..7db7455e 100644 --- a/src/content/questions/comp2804/2014-fall-final/16/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/16/solution.md @@ -1,12 +1,14 @@ -$ Pr(A|B) = \frac{|A \cap B|}{|B|} $ + +$ Pr(A|B) = \frac{|A \cap B|}{|B|} $ + $ Pr(A|B) = \frac{1}{3} $ diff --git a/src/content/questions/comp2804/2014-fall-final/17/solution.md b/src/content/questions/comp2804/2014-fall-final/17/solution.md index 3ad13f60..77dd6786 100644 --- a/src/content/questions/comp2804/2014-fall-final/17/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/17/solution.md @@ -1,17 +1,21 @@ -$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $ -$ \frac{1}{5} = \frac{1}{2} \cdot \frac{1}{2} $ -$ \frac{1}{5} = \frac{1}{4} $ + +$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $ + +$ \frac{1}{5} = \frac{1}{2} \cdot \frac{1}{2} $ + +$ \frac{1}{5} = \frac{1}{4} $ + Because the equation is false, these events are not independent. diff --git a/src/content/questions/comp2804/2014-fall-final/18/solution.md b/src/content/questions/comp2804/2014-fall-final/18/solution.md index 2b2273dc..c373cb4a 100644 --- a/src/content/questions/comp2804/2014-fall-final/18/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/18/solution.md @@ -1,17 +1,21 @@ -$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $ -$ \frac{3}{10} = \frac{1}{2} \cdot \frac{3}{5} $ -$ \frac{3}{10} = \frac{3}{10} $ + +$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $ + +$ \frac{3}{10} = \frac{1}{2} \cdot \frac{3}{5} $ + +$ \frac{3}{10} = \frac{3}{10} $ + Because the equation is true, these events are independent. diff --git a/src/content/questions/comp2804/2014-fall-final/19/solution.md b/src/content/questions/comp2804/2014-fall-final/19/solution.md index f7efe57a..9ad9957d 100644 --- a/src/content/questions/comp2804/2014-fall-final/19/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/19/solution.md @@ -1,17 +1,21 @@ We need to determine if events $ A $ and $ B $ are independent. -Now, let's check if $ Pr(A) \cdot Pr(B) = Pr(A \cap B) $ -$ \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} $ -$ \frac{1}{8} = \frac{1}{8} $ + +Now, let's check if $ Pr(A) \cdot Pr(B) = Pr(A \cap B) $ + +$ \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} $ + +$ \frac{1}{8} = \frac{1}{8} $ + The events $ A $ and $ B $ are independent. Thus, the statement is true. diff --git a/src/content/questions/comp2804/2014-fall-final/2/solution.md b/src/content/questions/comp2804/2014-fall-final/2/solution.md index 4c20b45b..e99d268d 100644 --- a/src/content/questions/comp2804/2014-fall-final/2/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/2/solution.md @@ -1,17 +1,19 @@ -$ |A| + |B| + |C| = 5^{13} + \binom{13}{1} \cdot 5 \cdot 5^{12} + \binom{13}{2} \cdot 5^2 \cdot 5^{11} $ + +$ |A| + |B| + |C| = 5^{13} + \binom{13}{1} \cdot 5 \cdot 5^{12} + \binom{13}{2} \cdot 5^2 \cdot 5^{11} $ + $ |A| + |B| + |C| = 5^{13} + 13 \cdot 5^{13} + \binom{13}{2} \cdot 5^{13} $ diff --git a/src/content/questions/comp2804/2014-fall-final/20/solution.md b/src/content/questions/comp2804/2014-fall-final/20/solution.md index e69de29b..a56c0762 100644 --- a/src/content/questions/comp2804/2014-fall-final/20/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/20/solution.md @@ -0,0 +1,3 @@ +I mean, there are a total of 4 options, and only one of them is right. + +The probability of getting the right answer is $ \frac{1}{4} $. diff --git a/src/content/questions/comp2804/2014-fall-final/21/solution.md b/src/content/questions/comp2804/2014-fall-final/21/solution.md index 1c37d2d8..23ff3909 100644 --- a/src/content/questions/comp2804/2014-fall-final/21/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/21/solution.md @@ -1,48 +1,43 @@ -subsection\*{Step-by-Step Solution} - -1. **Possible Outcomes and Corresponding Values:** +1. **Possible Outcomes and Corresponding Values:**
There are four possible outcomes when flipping two fair coins: - begin{itemize} -
  • First coin is heads (H), second coin is heads (H): $(H, H)$ -
  • First coin is heads (H), second coin is tails (T): $(H, T)$ -
  • First coin is tails (T), second coin is heads (H): $(T, H)$ -
  • First coin is tails (T), second coin is tails (T): $(T, T)$ - end{itemize} + -2. **Winning Amount for Each Outcome:** +2. **Winning Amount for Each Outcome:**
    For each outcome, we determine the amount $ X $ that you win: - begin{itemize} -
  • $(H, H)$: The first coin is heads, so we don't lose. The second coin is heads, so we win 1 dollar. Thus, $ X = 1 $. -
  • $(H, T)$: The first coin is heads, so we don't lose. The second coin is tails, so we don't win. Thus, $ X = 0 $. -
  • $(T, H)$: The first coin is tails, so we lose (i.e., we win $-1$ dollars). The second coin is heads, so we win 1 dollar, but since the first coin being tails means we lose, this overrides the second coin's result. Thus, $ X = -1 $. -
  • $(T, T)$: The first coin is tails, so we lose (i.e., we win $-1$ dollars). The second coin is tails, so we don't win. Thus, $ X = -1 $. - end{itemize} + -3. **Calculate the Expected Value $ E(X) $:** -The expected value $ E(X) $ is calculated using the formula: -[ -E(X) = sum_{i} P(X = x_i) \cdot x_i -] -Each of the four outcomes has a probability of $ \frac{1}{4} $, as the coins are fair and independent. +3. **Calculate the Expected Value $ E(X) $:**
    +The expected value $ E(X) $ is calculated using the formula:
    +$E(X) = sum\_{i} P(X = x_i) \cdot x_i $
    +Each of the four outcomes has a probability of $ \frac{1}{4} $, as the coins are fair and independent.
    Therefore, we have: -begin{itemize} -
  • $ P(X = 1) = \frac{1}{4} $ (from the $(H, H)$ outcome) + -4. **Substitute the Values into the Expected Value Formula:** - [ - E(X) = 1 \cdot \frac{1}{4} + 0 \cdot \frac{1}{4} + (-1) \cdot \frac{1}{2} - ] - [ - E(X) = \frac{1}{4} + 0 - \frac{1}{2} - ] - [ - E(X) = \frac{1}{4} - \frac{2}{4} - ] - [ - E(X) = -\frac{1}{4} - ] +4. **Substitute the Values into the Expected Value Formula:**
    + $E(X) = 1 \cdot \frac{1}{4} + 0 \cdot \frac{1}{4} + (-1) \cdot \frac{1}{2}$
    + $E(X) = \frac{1}{4} + 0 - \frac{1}{2}$
    + $E(X) = \frac{1}{4} - \frac{2}{4}$
    + $E(X) = -\frac{1}{4}$ -Therefore, the expected value of $ X $ is $ boxed{-\frac{1}{4}} $. +Therefore, the expected value of $ X $ is $ \boxed{-\frac{1}{4}} $. diff --git a/src/content/questions/comp2804/2014-fall-final/22/solution.md b/src/content/questions/comp2804/2014-fall-final/22/solution.md index 9f4b075e..bbb483b0 100644 --- a/src/content/questions/comp2804/2014-fall-final/22/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/22/solution.md @@ -1,28 +1,23 @@ -subsection\*{Step-by-Step Solution} +1. **Probability of Getting $ HTH $:**
    + The probability of getting a head followed by a tail followed by a head in three consecutive flips is:
    + $P(HTH) = P(H) \times P(T) \times P(H) = \left( \frac{1}{2}\right) \times \left( \frac{1}{2}\right) \times \left( \frac{1}{2}\right) = \frac{1}{8} $ -1. **Probability of Getting $ HTH $:** - The probability of getting a head followed by a tail followed by a head in three consecutive flips is: - [ - P(HTH) = P(H) times P(T) times P(H) = left( \frac{1}{2}right) times left( \frac{1}{2}right) times left( \frac{1}{2}right) = \frac{1}{8} - ] -2. **Number of Possible $ HTH $ in 6 Flips:** - Since we are looking for consecutive $ HTH $ patterns, we need to consider overlapping patterns in the sequence. - In a sequence of 6 coin flips, the positions where a $ HTH $ can start are: - begin{itemize} -
  • 1st position: $ {1, 2, 3} $ -
  • 2nd position: $ {2, 3, 4} $ -
  • 3rd position: $ {3, 4, 5} $ -
  • 4th position: $ {4, 5, 6} $ - end{itemize} +2. **Number of Possible $ HTH $ in 6 Flips:**
    + Since we are looking for consecutive $ HTH $ patterns, we need to consider overlapping patterns in the sequence.
    + In a sequence of 6 coin flips, the positions where a $ HTH $ can start are:
    + Therefore, there are 4 possible positions for a $ HTH $ to occur. -3. **Expected Value Contribution from Each Position:** - For each position, the expected value contribution is: - [ - E(text{one } HTH) = P(HTH) times text{payout} = \frac{1}{8} times 5 = \frac{5}{8} - ] -4. **Total Expected Value $ E(X) $:** - Since there are 4 possible positions, the total expected value is the sum of the contributions from each position: - [ - E(X) = 4 times \frac{5}{8} = \frac{20}{8} = \frac{5}{2} - ] - Therefore, the expected value of $ X $ is $ boxed{ \frac{5}{2}} $. + +3. **Expected Value Contribution from Each Position:**
    + For each position, the expected value contribution is:
    + $E(\text{one } HTH) = P(HTH) \times \text{payout} = \frac{1}{8} \times 5 = \frac{5}{8}$
    + +4. **Total Expected Value $ E(X) $:**
    + Since there are 4 possible positions, the total expected value is the sum of the contributions from each position:
    + $E(X) = 4 \times \frac{5}{8} = \frac{20}{8} = \frac{5}{2}$
    + Therefore, the expected value of $ X $ is $ \boxed{ \frac{5}{2}} $. diff --git a/src/content/questions/comp2804/2014-fall-final/23/solution.md b/src/content/questions/comp2804/2014-fall-final/23/solution.md index 358a31d4..143301ea 100644 --- a/src/content/questions/comp2804/2014-fall-final/23/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/23/solution.md @@ -1,6 +1,11 @@ So um, this algorithm pretty much counts the number of attempts we make until we get $HHH$ or $TTT$ + We have a $ \frac{1}{8} $ chance of flipping an $HHH$ + We have a $ \frac{1}{8} $ chance of flipping an $TTT$ + This gives us a $ \frac{1}{4} $ chance of flipping either $HHH$ or $TTT$ + By the geometric distribution, the expected value of the number of attempts until we get a success is $ \frac{1}{p} $ + Therefore, the expected value of $X$ is $ \frac{1}{ \frac{1}{4}} = 4 $ diff --git a/src/content/questions/comp2804/2014-fall-final/24/solution.md b/src/content/questions/comp2804/2014-fall-final/24/solution.md index 06e511e6..ab15cfb1 100644 --- a/src/content/questions/comp2804/2014-fall-final/24/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/24/solution.md @@ -1,12 +1,18 @@ -$ \mathbb{E}(X) = 2 \cdot \mathbb{E}(\text{number of heads}) - 3 \cdot \mathbb{E}(\text{number of tails}) $ -$ \mathbb{E}(X) = 2 \cdot \sum*{i=0}^{n} Pr(A) - 3 \cdot \sum*{i=0}^{n} Pr(B) $ -$ \mathbb{E}(X) = 2 \cdot n \cdot \frac{1}{2} - 3 \cdot n \cdot \frac{1}{2} $ -$ \mathbb{E}(X) = n - \frac{3}{2}n $ -$ \mathbb{E}(X) = \frac{2n}{2} - \frac{3n}{2} $ + +$ \mathbb{E}(X) = 2 \cdot \mathbb{E}(\text{number of heads}) - 3 \cdot \mathbb{E}(\text{number of tails}) $ + +$ \mathbb{E}(X) = 2 \cdot \sum*{i=0}^{n} Pr(A) - 3 \cdot \sum*{i=0}^{n} Pr(B) $ + +$ \mathbb{E}(X) = 2 \cdot n \cdot \frac{1}{2} - 3 \cdot n \cdot \frac{1}{2} $ + +$ \mathbb{E}(X) = n - \frac{3}{2}n $ + +$ \mathbb{E}(X) = \frac{2n}{2} - \frac{3n}{2} $ + $ \mathbb{E}(X) = \frac{-n}{2} $ diff --git a/src/content/questions/comp2804/2014-fall-final/3/solution.md b/src/content/questions/comp2804/2014-fall-final/3/solution.md index e69de29b..9d3244b2 100644 --- a/src/content/questions/comp2804/2014-fall-final/3/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/3/solution.md @@ -0,0 +1,9 @@ +$ \sum\_{k=0}^{45} \binom{45}{k} {{(-3)}^{2k}} $ + +$ = \sum\_{k=0}^{45} \binom{45}{k} {9}^{k} $ + +$ = \sum\_{k=0}^{45} \binom{45}{k} {9}^{k} {1}^{45-k} $ + +$= {(1+9)}^{45} $ + +$= {10}^{45} $ diff --git a/src/content/questions/comp2804/2014-fall-final/4/solution.md b/src/content/questions/comp2804/2014-fall-final/4/solution.md index fe6f9f85..03d20baf 100644 --- a/src/content/questions/comp2804/2014-fall-final/4/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/4/solution.md @@ -1,19 +1,24 @@ -$ |A \cup B| = |A| + |B| - |A \cap B| $ -$ |A \cup B| = 2^{10} + 2^{11} - 2^8 $ -$ |A \cup B| = 1 \cdot 2^{10} + 2 \cdot 2^{10} - 2^8$ -$ |A \cup B| = (1+2) \cdot 2^{10} - 2^8 $ + +$ |A \cup B| = |A| + |B| - |A \cap B| $ + +$ |A \cup B| = 2^{10} + 2^{11} - 2^8 $ + +$ |A \cup B| = 1 \cdot 2^{10} + 2 \cdot 2^{10} - 2^8$ + +$ |A \cup B| = (1+2) \cdot 2^{10} - 2^8 $ + $ |A \cup B| = 3 \cdot 2^{10} - 2^8 $ diff --git a/src/content/questions/comp2804/2014-fall-final/5/solution.md b/src/content/questions/comp2804/2014-fall-final/5/solution.md index eb01dfaa..b8ec5c2d 100644 --- a/src/content/questions/comp2804/2014-fall-final/5/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/5/solution.md @@ -4,4 +4,4 @@ Let's write out all recursive cases for $S_n$
  • $0, S_{n-1} $
  • $1, 0, S_{n-2} $ -$S*n = S*{n-1} + S_{n-2} $ +$S_n = S_{n-1} + S_{n-2} $ diff --git a/src/content/questions/comp2804/2014-fall-final/6/solution.md b/src/content/questions/comp2804/2014-fall-final/6/solution.md index 7e50e8db..58caea3a 100644 --- a/src/content/questions/comp2804/2014-fall-final/6/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/6/solution.md @@ -1,10 +1,19 @@ A = Like 8:30am classes + $ |A| = 40 $ + B = Like COMP 2804 + $ |B| = 30 $ + $ |\overline{A \cup B}| = 50 $ + $ |A \cup B| = 100-50 $ + $ |A \cup B| = 50 $ + $ |A \cap B| = |A| + |B| - |A \cup B| $ + $ |A \cap B| = 40 + 30 - 50 $ + $ |A \cap B| = 20 $ diff --git a/src/content/questions/comp2804/2014-fall-final/7/solution.md b/src/content/questions/comp2804/2014-fall-final/7/solution.md index c89f5f32..eaf75d5d 100644 --- a/src/content/questions/comp2804/2014-fall-final/7/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/7/solution.md @@ -1,9 +1,17 @@ A = Event that we pick 7 blue balls + First, we choose 7 of the m blue balls: $ \binom{m}{7} $ + The first ball has 7 choices, the second ball has 6 choices, the third ball has 5 choices, and so on until the seventh ball has 1 choice: $ 7! $ + $ |A| = \binom{m}{7} \cdot 7! $ + B = Event that we pick 7 red balls + First, we choose 7 of the n red balls: $ \binom{n}{7} $ + The first ball has 7 choices, the second ball has 6 choices, the third ball has 5 choices, and so on until the seventh ball has 1 choice: $ 7! $ + $ |B| = \binom{n}{7} \cdot 7! $ + $ |A| + |B| = \binom{m}{7} \cdot 7! + \binom{n}{7} \cdot 7! $ diff --git a/src/content/questions/comp2804/2014-fall-final/8/solution.md b/src/content/questions/comp2804/2014-fall-final/8/solution.md index 7636cbb4..489a4ff4 100644 --- a/src/content/questions/comp2804/2014-fall-final/8/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/8/solution.md @@ -1,9 +1,13 @@ -$ \binom{10}{3} \cdot \binom{7}{2} \cdot \binom{5}{5} $ -$ \frac{10!}{3!7!} \cdot \frac{7!}{2!5!} \cdot \frac{5!}{5!0!} $ -$ \frac{10!}{3!1} \cdot \frac{1}{2!5!} \cdot \frac{1}{1 \cdot 1} $ -$ \frac{10!}{3!2!5!} $ + +$ \binom{10}{3} \cdot \binom{7}{2} \cdot \binom{5}{5} $ + +$ = \frac{10!}{3!7!} \cdot \frac{7!}{2!5!} \cdot \frac{5!}{5!0!} $ + +$ = \frac{10!}{3!1} \cdot \frac{1}{2!5!} \cdot \frac{1}{1 \cdot 1} $ + +$ = \frac{10!}{3!2!5!} $ diff --git a/src/content/questions/comp2804/2014-fall-final/9/solution.md b/src/content/questions/comp2804/2014-fall-final/9/solution.md index 8a5c0f4a..335fffc0 100644 --- a/src/content/questions/comp2804/2014-fall-final/9/solution.md +++ b/src/content/questions/comp2804/2014-fall-final/9/solution.md @@ -1,9 +1,11 @@ We can solve this using stars and bars approach + There are 13 stars and 2 extra positions to place the bars/dividers between rows of stars + As a result, we pick 2 positions out of 13 and 2 positions to place the bars: $ \binom{15}{2} $