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?
  
<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>
  
[[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>.
  
== Solutions ==
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==See Also==
We have 4 choices for desserts.
 
  
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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{{AMC10 box|year=2012|ab=B|num-b=10|num-a=12}}
It is also given that there must be cake on Friday.
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{{MAA Notice}}
 
 
So,
 
 
 
<math>4*3*3*3*3*1*3=\boxed{972}</math>
 
 
 
OR
 
 
 
 
 
<math> \textbf{(B)}</math>
 

Revision as of 23:27, 28 October 2015

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 $\boxed{\textbf{(A)}\ 729}$.

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
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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