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Bunuel
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Why isn't the answer E? I am confused...
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(A) 30 as 3C1.4C3 + 3C2.4C2, rest all other combinations do not satisfy the minimum quantity requirement.

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7!/3!4!=30

Ans A
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daniformic
7!/3!4!=30

Ans A

you are wrong: 7!/3!4!=35
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Bunuel
A question paper consists of two sections A and B having respectively 3 and 4 questions. Four questions are to be solved to qualify in that paper. It is compulsory to solve at least one question from Section A and at least two questions from Section B. In how many ways can a candidate select the questions to qualify in that paper?


(A) 30
(B) 18
(C) 48
(D) 60
(E) 72


Case 1: 1 question correct in A, 3 questions correct in B
From the 3 questions in A, the number of ways to choose 1 to be correct = 3C1 = 3
From the 4 questions in B, the number of ways to choose 3 to be correct = 4C3 \(= \frac{4*3*2}{3*2*1} = 4\)
To combine these options, we multiply:
3*4 = 12

Case 2: 2 questions correct in A, 2 questions correct in B
From the 3 questions in A, the number of ways to choose 2 to be correct = 3C2 \(=\frac{3*2}{2*1} = 3\)
From the 4 questions in B, the number of ways to choose 2 to be correct = 4C2 \(= \frac{4*3}{2*1} = 6\)
To combine these options, we multiply:
3*6 = 18

Total number of ways = Case 1 + Case 2 = 12+18 = 30

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3c2×4c2+ 3c1+4c3= 30

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