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> | + | <math> \textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304 </math> |
| − | + | == Solution == | |
| − | <math>\boxed{\textbf{(A)}\ 729}</math> | + | Desserts must be chosen for <math>7</math> days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday. |
| + | |||
| + | 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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| + | == Solution 2== | ||
| + | |||
| + | 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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| + | - Sliced_Bread | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AMC10 box|year=2012|ab=B|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:08, 22 September 2023
Contents
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?
Solution
Desserts must be chosen for 
days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.
There are 
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 choices for dessert on Thursday. Once dessert for Thursday is selected, there are choices for dessert on Wednesday, once Wednesday's dessert is selected there are choices for dessert on Tuesday, etc. Thus, there are choices for dessert for each of 
days, so the total number of possible dessert menus is 
, or 
.
Solution 2
There are 
ways for the desserts to be chosen. By symmetry, any of the desserts that are chosen on Friday share 
of the total arrangements. Therefore our answer is 
- 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
