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Difference between revisions of "2012 AMC 10B Problems/Problem 11"

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A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
 
A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
  
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<math> \textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\2304 </math>
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<math> \textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304 </math>
  
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[[2012 AMC 10B Problems/Problem 11|Solution]]
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== Solution ==
  
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Desserts must be chosen for <math>7</math> days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.
  
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There are <math>3</math> choices for dessert on Saturday: pie, ice cream, or pudding, as there must be cake on Friday and the same dessert may not be served two days in a row. Likewise, there are <math>3</math> choices for dessert on Thursday. Once dessert for Thursday is selected, there are <math>3</math> choices for dessert on Wednesday, once Wednesday's dessert is selected there are <math>3</math> choices for dessert on Tuesday, etc. Thus, there are <math>3</math> choices for dessert for each of <math>6</math> days, so the total number of possible dessert menus is <math>3^6</math>, or <math>\boxed{\textbf{(A)}\ 729}</math>.
  
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== Solutions ==
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== Solution 2==
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We have 4 choices for desserts.
 
  
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However, the same dessert cannot be served for 2 straight days, meaning that you only have 3 choices for a dessert for the next day.  
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There are <math>4 \cdot 3^6</math> ways for the desserts to be chosen. By symmetry, any of the desserts that are chosen on Friday share <math>\frac{1}{4}</math> of the total arrangements. Therefore our answer is <math>\frac{4\cdot3^6}{4} = 3^6 = \boxed{729}.</math>
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It is also given that there must be cake on Friday.
 
  
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So,
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- Sliced_Bread
  
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<math>4*3*3*3*3*1*3=\boxed{972}</math>
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==See Also==
  
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OR
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{{AMC10 box|year=2012|ab=B|num-b=10|num-a=12}}
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{{MAA Notice}}
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<math> \textbf{(B)}</math>
 

Latest revision as of 18:08, 22 September 2023

Problem 11

A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?

$\textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304$

Solution

Desserts must be chosen for $7$ days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.

There are $3$ choices for dessert on Saturday: pie, ice cream, or pudding, as there must be cake on Friday and the same dessert may not be served two days in a row. Likewise, there are $3$ choices for dessert on Thursday. Once dessert for Thursday is selected, there are $3$ choices for dessert on Wednesday, once Wednesday's dessert is selected there are $3$ choices for dessert on Tuesday, etc. Thus, there are $3$ choices for dessert for each of $6$ days, so the total number of possible dessert menus is $3^6$, or $\boxed{\textbf{(A)}\ 729}$.

Solution 2

There are $4 \cdot 3^6$ ways for the desserts to be chosen. By symmetry, any of the desserts that are chosen on Friday share $\frac{1}{4}$ of the total arrangements. Therefore our answer is $\frac{4\cdot3^6}{4} = 3^6 = \boxed{729}.$

- Sliced_Bread

See Also

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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