binomial theorem solution fixes + tag fixes

src/content/questions/comp2804/2014-fall-midterm/7/solution.md

@@ -1,11 +1,11 @@
 ${(5x-36)}^{100}$
 
-$=\sum^{100}_{k=0} \binom{100}{k}{(5x)}^k {(-3y)}^{n-k}$
+$ = \sum^{100}_{k=0} \binom{100}{k} {(5x)}^{n-k} {(-3y)}^{k} $
 
-$=\binom{100}{20}{(5x)}^{20} {(-3y)}^{80}$
+We only consider $k=80$, as it results in $y^{80}$.
 
-$=\binom{100}{20}5^{20} 3^{80} x^{20} y^{80}$
+$ = \binom{100}{80} \cdot {(5x)}^{100-80} \cdot {(-3y)}^{80} $
 
-$=\binom{100}{80}5^{20} 3^{80} x^{20} y^{80}$
+$ = \binom{100}{80} \cdot 5^{20} \cdot {(-3)}^{80} \cdot x^{20} \cdot y^{80} $
 
-$=\binom{100}{80}5^{20} 3^{80}$ (this is the coefficient)
+$ = \binom{100}{80} \cdot 5^{20} \cdot 3^{80} $ (final answer, i.e. the coefficient of $x^{20} y^{80}$)

src/content/questions/comp2804/2015-fall-final/4/solution.md

@@ -1,7 +1,11 @@
-$ = \sum^{20}\_{k=0} \binom{20}{k} {(-3x)}^{k} {(5y)}^{20-k} $
+$ = \sum^{20}_{k=0} \binom{20}{k} {(-3x)}^{n-k} {(5y)}^{k} $
 
-$ = \binom{20}{15} {(-3)}^{15} {5}^{5} x^{15} y^5 $
+We only consider $k=5$, as it results in $y^{5}$.
 
-$ = - \binom{20}{15} {(3)}^{15} {5}^{5} x^{15} y^5 $
+$ = \binom{20}{5} \cdot {(-3x)}^{20-5} \cdot {(5y)}^{5} $
 
-Thus, the coefficient of $x^{15}y^{5}$ in the expansion of ${(-3x + 5y)}^{20}$ is $ - \binom{20}{15} {(3)}^{15} {5}^{5} $
+$ = \binom{20}{5} \cdot {(-3)}^{15} \cdot {5}^{5} \cdot x^{15} \cdot y^5 $
+
+$ = - \binom{20}{5} \cdot {3}^{15} \cdot {5}^{5} \cdot x^{15} \cdot y^5 $
+
+Thus, the coefficient of $x^{15}y^{5}$ in the expansion of ${(-3x + 5y)}^{20}$ is $ - \binom{20}{5} \cdot {3}^{15} \cdot {5}^{5} $

src/content/questions/comp2804/2015-fall-midterm/7/index.md

@@ -8,5 +8,5 @@ solution: comp2804/2015-fall-midterm/7/solution.md
 tags:
   - comp2804
   - comp2804-midterm
-  - comp2804-newton's-binomial-theorem
+  - comp2804-the-pigeonhole-principle
 ---

src/content/questions/comp2804/2015-fall-midterm/8/index.md

@@ -8,5 +8,5 @@ solution: comp2804/2015-fall-midterm/8/solution.md
 tags:
   - comp2804
   - comp2804-midterm
-  - comp2804-counting-solutions-of-linear-equations
+  - comp2804-newton's-binomial-theorem
 ---

src/content/questions/comp2804/2015-fall-midterm/8/solution.md

@@ -1,4 +1,4 @@
-$=\sum^{88}_{k=0} \binom{88}{k}{(3x)}^k {(-17y)}^{88-k}$
+$=\sum^{88}_{k=0} \binom{88}{k}{(3x)}^{88-k} {(-17y)}^{k}$
 
 $=\binom{88}{7}{(3x)}^{81} {(-17y)}^{7}$
 

src/content/questions/comp2804/2015-winter-final/4/solution.md

@@ -1,7 +1,13 @@
 $ {(2x-7y)}^{15} $
 
-$= \sum\_{k=4}^{15} \binom{15}{k} {(2x)}^{k} {(-7y)}^{15-k} $
+$ = \sum_{k=0}^{15} \binom{15}{k} {(2x)}^{n-k} {(-7y)}^{k} $
 
-$ = \binom{15}{4} 2^{4} {(-7)}^{11} x^4 y^{11}$
+We only consider $k=11$, as it results in $y^{11}$.
 
-$ = - \binom{15}{4} 2^{4} {(7)}^{11} $
+$ = \binom{15}{11} \cdot {(2x)}^{15-11} \cdot {(-7y)}^{11} $
+
+$ = \binom{15}{11} \cdot 2^{4} \cdot {(-7)}^{11} \cdot x^4 \cdot y^{11} $
+
+$ = - \binom{15}{4} \cdot 2^{4} \cdot 7^{11} \cdot x^4 \cdot y^{11} $
+
+Thus, the coefficient of $ x^{4}y^{11} $ in the expansion of $ {(2x-7y)}^{15} $ is $ - \binom{15}{11} \cdot {2}^{4} \cdot {7}^{11} $

src/content/questions/comp2804/2016-fall-midterm/9/solution.md

@@ -1,7 +1,11 @@
-$ = \sum^{50}\_{k=0} \binom{50}{k} {(5x)}^{50-k} {(-7y)}^k $
+$ = \sum^{50}_{k=0} \binom{50}{k} {(5x)}^{n-k} {(-7y)}^k $
 
-$ = \sum^{50}\_{k=0} \binom{50}{26} {(5x)}^{50-26} {(-7y)}^k $
+We only consider $k=26$, as it results in $y^{26}$.
 
-$ = \binom{50}{24} {(5)}^{24} x^{24} {(-7)}^{26} y^{26} $
+$ = \binom{50}{26} \cdot {(5x)}^{50-26} \cdot {(-7y)}^{26} $
 
-$ = \binom{50}{26} {(5)}^{24} {(-7)}^{26} x^{24} y^{26} $
+$ = \binom{50}{26} \cdot {(5)}^{24} \cdot x^{24} \cdot {(-7)}^{26} \cdot y^{26} $
+
+$ = \binom{50}{26} \cdot 5^{24} \cdot 7^{26} \cdot x^{24} \cdot y^{26} $
+
+Thus, the coefficient is $ \binom{50}{26} \cdot 5^{24} \cdot 7^{26} $

src/content/questions/comp2804/2017-fall-midterm/9/solution.md

@@ -1,11 +1,11 @@
-$ = \sum^{100}\_{k=0} \binom{100}{k} {(7x)}^{k} {(-13y)}^{100-k} $
+$ = \sum^{100}_{k=0} \binom{100}{k} {(7x)}^{n-k} {(-13y)}^{k} $
 
-$ = \sum^{100}\_{k=0} \binom{100}{20} {(7x)}^{20} {(-13y)}^{100-20} $
+We only consider $k=80$, as it results in $y^{80}$.
 
-$ = \binom{100}{20} {(7)}^{20} x^{20} {(-13)}^{80} y^{80} $
+$ = \binom{100}{80} \cdot {(7x)}^{100-80} \cdot {(-13y)}^{80} $
 
-$ = \binom{100}{20} {(7)}^{20} {(-13)}^{80} x^{20} y^{80} $
+$ = \binom{100}{80} \cdot {(7)}^{20} \cdot x^{20} \cdot {(-13)}^{80} \cdot y^{80} $
 
-$ = \binom{100}{20} {(7)}^{20} {(13)}^{80} x^{20} y^{80} $
+$ = \binom{100}{80} \cdot 7^{20} \cdot 13^{80} \cdot x^{20} \cdot y^{80} $
 
-Thus, the coefficient is $ \binom{100}{20} {(7)}^{20} {(13)}^{80} $
+Thus, the coefficient is $ \binom{100}{80} \cdot 7^{20} \cdot 13^{80} $

src/content/questions/comp2804/2018-fall-final/6/solution.md

@@ -1,7 +1,11 @@
-$ = \sum\_{k = 35}^{55} \binom{55}{k} {(5x)}^{k} {(-3y)}^{n-k} $
+$ = \sum_{k=0}^{55} \binom{55}{k} {(5x)}^{n-k} {(-3y)}^{k} $
 
-$ = \binom{55}{20} {(5x)}^{20} {(-3y)}^{35} $
+We only consider $k=35$, as it results in $y^{35}$.
 
-$ = - \binom{55}{20} 5^{20} 3^{35} x^{20} y^{35}$
+$ = \binom{55}{35} \cdot {(5x)}^{55-35} \cdot {(-3y)}^{35} $
 
-The coefficient is $ - \binom{55}{20} 5^{20} 3^{35} $
+$ = \binom{55}{35} \cdot 5^{20} \cdot {(-3)}^{35} \cdot x^{20} \cdot y^{35}$
+
+$ = - \binom{55}{35} \cdot 5^{20} \cdot 3^{35} \cdot x^{20} \cdot y^{35}$
+
+The coefficient is $ - \binom{55}{35} \cdot 5^{20} \cdot 3^{35} $

src/content/questions/comp2804/2022-winter-final/6/solution.md

@@ -1,13 +1,15 @@
 $ (2x - 3y)^{30} $
 
-$ = \sum\_{k=10}^{30} \binom{30}{k} \cdot (2x)^{k} \cdot (-3y)^{30-k} $
+$ = \sum_{k=0}^{30} \binom{30}{k} \cdot (2x)^{n-k} \cdot (-3y)^{k} $
 
-$ = \binom{30}{10} \cdot (2x)^{10} \cdot (-3y)^{30-10} $
+We only consider $k=20$, as it results in $y^{20}$.
 
-$ = \binom{30}{10} \cdot (2x)^{10} \cdot (-3y)^{20} $
+$ = \binom{30}{20} \cdot (2x)^{30-20} \cdot (-3y)^{20} $
 
-$ = \binom{30}{10} \cdot 2^{10} \cdot (-3)^{20} \cdot x^{10} \cdot y^{20} $
+$ = \binom{30}{20} \cdot (2x)^{10} \cdot (-3y)^{20} $
 
-$ = \binom{30}{10} \cdot 2^{10} \cdot (3)^{20} \cdot x^{10} \cdot y^{20} $
+$ = \binom{30}{20} \cdot 2^{10} \cdot (-3)^{20} \cdot x^{10} \cdot y^{20} $
 
-From this equation, we can see that the coefficient (aka the real numbers) are: $\binom{30}{10} \cdot 2^{10} \cdot (3)^{20}$
+$ = \binom{30}{20} \cdot 2^{10} \cdot (3)^{20} \cdot x^{10} \cdot y^{20} $
+
+From this equation, we can see that the coefficient (aka the real numbers) are: $\binom{30}{20} \cdot 2^{10} \cdot 3^{20}$