Difference between revisions of "2024 AMC 10A Problems/Problem 8"

Revision as of 21:27, 20 March 2025

Problem

In how many ways can $6$ juniors and $6$ seniors form $3$ disjoint teams of $4$ people so that each team has $2$ juniors and $2$ seniors?

$\textbf{(A)}~720\qquad\textbf{(B)}~1350\qquad\textbf{(C)}~2700\qquad\textbf{(D)}~3280\qquad\textbf{(E)}~8100$

Solution 1

The number of ways in which we can choose the juniors for the team are ${6\choose2}{4\choose2}{2\choose2}=15\cdot6\cdot1=90$ . Similarly, the number of ways to choose the seniors are the same, so the total is $90\cdot90=8100$ . But we must divide the number of permutations of three teams, since the order in which the teams were chosen never mattered, which is $3!$ . Thus the answer is $\frac{8100}{3!}=\frac{8100}{6}=\boxed{\textbf{(B) }1350}$ !

~eevee9406 ~small edits by NSAoPS

Solution 2

Let's first suppose that the three teams are distinguishable. The number of ways to choose the first team of two juniors and two seniors is ${6\choose2}{6\choose2} = 15\cdot 15 = 225$ . Now, only four juniors and four seniors are left to choose the second team. Thus, the second team can be formed in ${4\choose2}{4\choose2} = 6\cdot 6 = 36$ ways. There are now only two juniors and two seniors left, so the third team can only be formed in one way. Thus, the total number of ways in which we can choose two juniors and two teams for three distinguishable teams is $225\cdot 36 = 8100$ ways. However, the problem does not require the teams to be distinguishable. Therefore, we must divide by $3!=6$ to find the answer for three indistinguishable teams.

The answer is $\frac{8100}{6} =\boxed{\textbf{(B) }1350}$

~jjjxi

~edit by yanes04 (wrong number)

Video Solution by Pi Academy

https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv

Video Solution 1 by Power Solve

https://youtu.be/j-37jvqzhrg?si=IBSPzNSvdIodGvZ7&t=1145

Video Solution by Daily Dose of Math

https://youtu.be/AEd5tf1PJxk

~Thesmartgreekmathdude

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=_o5zagJVe1U

@@ Line 1: / Line 1: @@
-== Problem ==
+==Problem==
+In how many ways can <math>6</math> juniors and <math>6</math> seniors form <math>3</math> disjoint teams of <math>4</math> people so that each team has <math>2</math> juniors and <math>2</math> seniors?
-Amy, Bomani, Charlie, and Daria work in a chocolate factory. On Monday Amy, Bomani, and Charlie started working at <math>1:00 \ \mathrm{PM}</math> and were able to pack <math>4</math>, <math>3</math>, and <math>3</math> packages, respectively, every <math>3</math> minutes. At some later time, Daria joined the group, and Daria was able to pack <math>5</math> packages every <math>4</math> minutes. Together, they finished packing <math>450</math> packages at exactly <math>2:45\ \mathrm{PM}</math>. At what time did Daria join the group?
+<math>\textbf{(A)}~720\qquad\textbf{(B)}~1350\qquad\textbf{(C)}~2700\qquad\textbf{(D)}~3280\qquad\textbf{(E)}~8100</math>
-<math>\textbf{(A) }1:25\text{ PM}\qquad\textbf{(B) }1:35\text{ PM}\qquad\textbf{(C) }1:45\text{ PM}\qquad\textbf{(D) }1:55\text{ PM}\qquad\textbf{(E) }2:05\text{ PM}</math>
+== Solution 1==
+The number of ways in which we can choose the juniors for the team are <math>{6\choose2}{4\choose2}{2\choose2}=15\cdot6\cdot1=90</math>. Similarly, the number of ways to choose the seniors are the same, so the total is <math>90\cdot90=8100</math>. But we must divide the number of permutations of three teams, since the order in which the teams were chosen never mattered, which is <math>3!</math>. Thus the answer is <math>\frac{8100}{3!}=\frac{8100}{6}=\boxed{\textbf{(B) }1350}</math>!
-== Solution 1 ==
+~eevee9406
+~small edits by NSAoPS
-Note that Amy, Bomani, and Charlie pack a total of <math>4+3+3=10</math> packages every <math>3</math> minutes.
+== Solution 2 ==
+Let's first suppose that the three teams are distinguishable. The number of ways to choose the first team of two juniors and two seniors is <math>{6\choose2}{6\choose2} = 15\cdot 15 = 225</math>. Now, only four juniors and four seniors are left to choose the second team. Thus, the second team can be formed in <math>{4\choose2}{4\choose2} = 6\cdot 6 = 36</math> ways. There are now only two juniors and two seniors left, so the third team can only be formed in one way. Thus, the total number of ways in which we can choose two juniors and two teams for three distinguishable teams is <math>225\cdot 36 = 8100</math> ways. However, the problem does not require the teams to be distinguishable. Therefore, we must divide by <math>3!=6</math> to find the answer for three indistinguishable teams.
-The total amount of time worked is <math>1</math> hour and <math>45</math> minutes, which when converted to minutes, is <math>105</math> minutes. This means that since Amy, Bomani, and Charlie worked for the entire <math>105</math> minutes, they in total packed <math>\dfrac{105}{3}\cdot10=350</math> packages.
+The answer is <math>\frac{8100}{6} =\boxed{\textbf{(B) }1350}</math>
-Since <math>450</math> packages were packed in total, then Daria must have packed <math>450-350=100</math> packages in total, and since she packs at a rate of <math>5</math> packages per <math>4</math> minutes, then Daria worked for <math>\dfrac{100}{5}\cdot4=80</math> minutes, therefore Daria joined <math>80</math> minutes before <math>2:45</math> PM, which was at <math>\boxed{\textbf{(A) }1:25\text{ PM}}</math>
+~jjjxi
-~Tacos_are_yummy_1, andliu766
+~edit by yanes04 (wrong number)
-== Solution 2 ==
+== Video Solution by Pi Academy ==
-Let the time, in minutes, elapsed between <math>1:00</math> and the time Daria joined the packaging be <math>x</math>. Since Amy packages <math>4</math> packages every <math>3</math> minutes, she packages <math>\frac{4}{3}</math> packages per minute. Similarly, we can see that both Bomani and Charlie package <math>1</math> package per minute, and Daria packages <math>\frac{5}{4}</math> packages every minute.
+https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
-Before Daria arrives, we can write the total packages packaged as <math>x(\frac{4}{3} + 1 + 1) = x(\frac{10}{3})</math>. Since there are <math>105</math> minutes between <math>1:00</math> and <math>2:45</math>, Daria works with the other three for <math>105-x</math> minutes, meaning for that time there are <math>(105-x)(\frac{4}{3} + 1 + 1 + \frac{5}{4}) = (105-x)(\frac{55}{12})</math> packages packaged.
+==Video Solution 1 by Power Solve ==
-Adding the two, we get <math>x(\frac{10}{3}) + (105-x)(\frac{55}{12}) = 450</math> (The total packaged in the entire time is <math>450</math>). Solving this equation, we get <math>x=25</math>, meaning Daria arrived <math>25</math> minutes after <math>1:00</math>, meaning the answer is <math>\boxed{\textbf{(A) }1:25\text{ PM}}</math>.
+https://youtu.be/j-37jvqzhrg?si=IBSPzNSvdIodGvZ7&t=1145
-~i_am_suk_at_math_2
+== Video Solution by Daily Dose of Math ==
-==Video Solution 1 by Power Solve ==
+https://youtu.be/AEd5tf1PJxk
-https://youtu.be/j-37jvqzhrg?si=bf4iiXH4E9NM65v8&t=996
+~Thesmartgreekmathdude
-== Video Solution 2 by Daily Dose of Math ==
+==Video Solution by SpreadTheMathLove==
-[//youtu.be/W5hES6aNXAk ~Thesmartgreekmathdude]
-==Video Solution 3 by SpreadTheMathLove==
 https://www.youtube.com/watch?v=_o5zagJVe1U
-== Video Solution 4 by Pi Academy ==
+==See also==
-https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
-== See Also ==
 {{AMC10 box|year=2024|ab=A|num-b=7|num-a=9}}
 {{MAA Notice}}
-[[Category:Introductory Algebra Problems]]

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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