Fritz is taking an examination that consists of two parts, A

Question

Fritz is taking an examination that consists of two parts, A and B, with the following instructions:

Part A contains three questions, and a student must answer two.
Part B contains four questions, and a student must answer two.
Part A must be completed before starting Part B.

In how many ways can the test be completed?

(A) 12
(B) 15
(C) 36
(D) 72
(E) 90

Answer is given 72. But I don't think so as the condition in bold is not fulfilled. Please explain.

Bunuel · Accepted Answer

No, that's not correct.

We use permutation for both parts because Fritz can answer the questions in each of them in any order. For example, in part A, he can answer the questions in the following order: {1st, 2nd}, {2nd, 1st}, {1st, 3rd}, {3rd, 1st}, {2nd, 3rd}, {3rd, 2nd}.

As for part A and part B. If we were not told that Part A must be completed before starting Part B, then the answer would be  2*(P^2_3*P^2_4) , meaning that in this case he could attempt part A first and then start part B or he could attempt part B first and then start part A.

Hope it's clear.

Bunuel · Answer

Since Part A must be completed before starting Part B, then Fritz must answer any two questions from three, in any order, from Part A and then any two questions from four, in any order, from Part B: P^2_3*P^2_4=6*12=72 . Answer: D. P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 2, and 3. Thank you.

justbequiet · Answer

Hi Bunuel,

Why is permutation used instead of combination for both operations when we can arrange the contents within part A & B in any order?

DesiGmat · Answer

Because it's mentioned in the question that Part A has to be complete before starting Part B. 
Order is important, hence Permutation.

Please correct me if i am wrong.

Many Thanks.

mvictor · Answer

picked E, since it was the only solution it popped into my mind:
started first with 3C2 - 3 ways to finish part A
then 4C2 = 6 ways to finish part B, but then, I got confused, since it should have been 3*6=18 ways.
nonetheless, this helped me to eliminate at least answer choice B :D

then I thought..hmm...
what if we have to split everything into stages, since we are told that A must be finished before B.
so we have to answer first question, then second. then move to B. at be, we have to answer first question, then second.
this can be done in:
A can thus be finished in: 3C1 = 3 and 2C2 = 2 ways, or 6 ways.
B can be finished in: 4C1 = 4 and 3C1 = 3 ways, or 12.
6*12 = 72 ways.

MT1988 · Answer

After reading the explanations i understand that as the question mentions ways to splve the paper the question is a permutation problem.
What i want to clarify further is that as Fritz must attempt 2 questions can he also attempt 3. the word only seems to be missing here.

Would like to understand what scenario we would take the possibility that he attempts 3 questions instead of the mandatory 2 into consideration.
words like only, minimum etc would help. would GMAT be more precise?

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Fritz is taking an examination that consists of two parts, A

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