A dessert chef prepares the dessert for every day of a week starting

I've started counting possibilities from Friday

S - 3 posibilities
M - 3
T - 3
W - 3
Th - 3
F - 1 (cake)
Sat - 3

As on Friday we have got a cake on Thursday and Saturday we can have only 3 possibilities (pudding, ice-cream, pie)
As on Thursday we have 3 possibilities, on Wednesday we will have 3 as well, not the one which was on Thursday and so on.

Cant bet my approach is 100%, any expert reply wont be redundant

IMO
Ans: A

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 ,

Starting with friday as 1, moving on to the rest of the days with 3.
3^6

Posted from my mobile device

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

Ugh it's soo annoying

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

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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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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