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)
- B)
- C)
- D)
- E)
Answer
A
Key insight
Fix Friday as cake first; every other day then has 3 options once its neighbor on the Friday side is chosen, giving 3^6.
Solution
Begin with the most constrained day. Friday is cake: choice.
Now move outward from Friday. Saturday must differ from Friday: choices. Thursday must differ from Friday: choices. Wednesday must differ from Thursday (whatever it is): choices. The same holds for Tuesday, Monday and Sunday, each differing from the day after it: choices apiece.
Every day except Friday contributes a factor of , so the number of menus is
The answer is .
Why this works
With a "no two consecutive equal" rule, each day has exactly one fewer option than the total, provided its neighbor is already decided. Starting at the fixed day and working outward keeps every step at choices; starting at Sunday leaves Thursday and Friday entangled. In general: place the forced element first, then build away from it.
Alternative approach
Without the birthday rule there are menus (Sunday free, each later day choices). By symmetry among the four desserts, Friday is cake in exactly one quarter of them: .
The trap
Counting forward from Sunday as 4 * 3^5 = 972 and then forcing Friday, which fails because Thursday may already be cake.
Common mistakes
- Counting forward from Sunday as 4 * 3^5 = 972 and then forcing Friday, which fails because Thursday may already be cake.
- Giving Saturday choices (forgetting it must differ from Friday's cake), which yields as well.
Techniques
Set up the equation/formula and compute; no special trick needed · Start from the end state / desired conclusion and reverse