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Bunuel
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) 729
(B) 972
(C) 1024
(D) 2187
(E) 2304

we will have 3 possibilities of dishes per day starting friday ;
total ; 3^6 ; 729
IMO A

Hi Archit3110 ,
Why not start with Sunday where we will have 4 possibilities and go on like
Monday - 3 , Tue-3, Wed-3, Thu- 2( 3-1 because cake on friday) , friday- 1(cake) , and saturday - 3

Could you explain why is this approach wrong?

TIA
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Archit3110
Bunuel
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) 729
(B) 972
(C) 1024
(D) 2187
(E) 2304

we will have 3 possibilities of dishes per day starting friday ;
total ; 3^6 ; 729
IMO A

Hi Archit3110 ,
Why not start with Sunday where we will have 4 possibilities and go on like
Monday - 3 , Tue-3, Wed-3, Thu- 2( 3-1 because cake on friday) , friday- 1(cake) , and saturday - 3

Could you explain why is this approach wrong?

TIA

neha283
well if you start from sunday with 4 options then how will be sure that the option for Friday viz. cake will be correct?
which is why its better to start from friday ,
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Starting with friday as 1, moving on to the rest of the days with 3.
3^6

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ugh got tricked again. I started on a Sunday so I had
4 * 3 * 3 *3*3*1*3

Ugh it's soo annoying
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Bunuel
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) 729
(B) 972
(C) 1024
(D) 2187
(E) 2304

We have only one option on Friday, which is a cake. Since the dessert on Saturday and Thursday cannot be a cake, there are 3 options for each of those days. The dessert on Wednesday must be different from Thursday; so there are 3 options for Wednesday as well. Similarly, the desert on Tuesday must be different from Wednesday; the desert on Monday must be different from Tuesday and the desert on Sunday must be different from Monday as well; therefore there are 3 options for each of those days as well. In total, there are 3^6 = 729 different menus.

Answer: A
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Hi ScottTargetTestPrep Bunuel

I'm slightly confused here. Your reply would be appreciated :)

Taking two cases here (Both starting from Saturday) with number of options to choose from and using one of them for reference,

Case 1:
Saturday: 3 options, let's say Pudding (Since Cake is not a possibility)
Sunday: 3 options, Ice Cream
Monday: 3 options, Pie
Tuesday: 3 options, Pudding
For Wednesday, let's say we got Pie here, this should imply that for Thursday we can not have Pie and Cake (We already have cake on Friday) leaving only 2 options to choose from.

Hence, it would be 3(Sat)x3(Sun)x3(Mon)x3(Tue)x3(Wed)x2(Thur)x1(Fri) = 3^5x2

Case 2: )
Only difference here is that on Wednesday, if we have cake instead of anything else, we are left with 3 options for Thursday
Saturday: 3 options, let's say Pudding (Since Cake is not a possibility)
Sunday: 3 options, Ice Cream
Monday: 3 options, Pie
Tuesday: 3 options, Pudding
For Wednesday, let's say we got Cake here, this means that for Thursday we can not have Cake, leaving us with 3 options to choose from

Hence, it would be 3(Sat)x3(Sun)x3(Mon)x3(Tue)x3(Wed)x3(Thur)x1(Fri) which is 3^6

Am i going wrong somewhere? Pls help!!

TIA
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Given: 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.

Asked: How many different dessert menus for the week are possible?

Number of different dessert menus = 1(Friday)*3*3*3*3*3*3 = 3^6 = 729

IMO A
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Hi ScottTargetTestPrep Bunuel

I'm slightly confused here. Your reply would be appreciated :)

Taking two cases here (Both starting from Saturday) with number of options to choose from and using one of them for reference,

Case 1:
Saturday: 3 options, let's say Pudding (Since Cake is not a possibility)
Sunday: 3 options, Ice Cream
Monday: 3 options, Pie
Tuesday: 3 options, Pudding
For Wednesday, let's say we got Pie here, this should imply that for Thursday we can not have Pie and Cake (We already have cake on Friday) leaving only 2 options to choose from.

Hence, it would be 3(Sat)x3(Sun)x3(Mon)x3(Tue)x3(Wed)x2(Thur)x1(Fri) = 3^5x2

Case 2: )
Only difference here is that on Wednesday, if we have cake instead of anything else, we are left with 3 options for Thursday
Saturday: 3 options, let's say Pudding (Since Cake is not a possibility)
Sunday: 3 options, Ice Cream
Monday: 3 options, Pie
Tuesday: 3 options, Pudding
For Wednesday, let's say we got Cake here, this means that for Thursday we can not have Cake, leaving us with 3 options to choose from

Hence, it would be 3(Sat)x3(Sun)x3(Mon)x3(Tue)x3(Wed)x3(Thur)x1(Fri) which is 3^6

Am i going wrong somewhere? Pls help!!

TIA

Your analysis of case 1 is correct; assuming the dessert on Wednesday is not a cake, there are 2 * 3^5 options. In the analysis of case 2, you assume that the dessert on Wednesday IS a cake, so there is only one option for that day. Thus, there are 3 * 3 * 3 * 3 * 1 * 3 * 1 = 3^5 options. In total, there are 2 * 3^5 + 3^5 = 3 * 3^5 = 3^6 options.
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