A teacher prepares a test. She gives 5 objective type questi

Question

A teacher prepares a test. She gives 5 objective type questions out of which 4 have to be answered. Find the total ways in which they can be answered if the first 2 questions have 3 choices and the last 3 have 4 choices.

A. 255
B. 816
C. 192
D. 100
E. 144

A. 255
B. 816
C. 192
D. 100
E. 144

AKProdigy87 · Answer

Two cases.

First Case: 2 Questions - 3 Choices, 2 Questions - 4 Choices

2C2 * 3C2 ways to select the questions
3C1 * 3C1 * 4C1 * 4C1 ways to answer the questions

= 1 * 3 * 3 * 3 * 4 * 4
= 27 * 16
= 432

Second Case: 1 Question - 3 Choices, 3 Questions - 4 Choices

2C1 * 3C3 ways to select the questions
3C1 * 4C1 * 4C1 * 4C1 ways to answer the questions

= 2 * 1 * 3 * 4 * 4 * 4
= 6 * 64
= 384

Total = 384 + 432
= 816

bhushan252 · Answer

Good one 'AKProdigy87' i could solve upto 
3 choices x 1 Q + 4 choices x 3 Q
3 choices x 2 Q + 4 choices x 2 Q

then i got all wrong...

Bunuel · Answer

So there are two types of questions: two with 3 choices and three with 4 choices.

In order to answer  exactly  4 out of 5 then either  both from the first type and any two from the second type  must be answered OR  any from the first type and all three from the second type .

First case:  (3*3)*(C^2_3*4*4)=432 ;
Second case  (C^1_2*3)*(4*4*4)=384 ;

So total:  432+384=816 .

Answer: B.

jlgdr · Answer

So we have 5 questions, pick 4 Only two ways to pick them are: 1 of the three choice questions and 3 of the 4 choice questions OR 2 of each First scenario (2C1)(3)(4^3) = 384 Second scenario (3C2)(3^2)(4^2) = 432 Add both = 816, or you know what don't even add them just note that UD has to be 6 only 1 answer choice that is b Hence answer is (B) Hope it helps Cheers! J

rever08 · Answer

First Case: 2 Questions with 3 Choices; 2 Questions with 4 Choices

2C2 * 3C2 ways to select the questions
and ways to answer = 3C1 * 3C1 * 4C1 * 4C1
= 1 * 3 * 3 * 3 * 4 * 4
= 27 * 16
= 432

Second Case: 1 Question with 3 Choices; 3 Questions with 4 Choices
2C1 * 3C3 ways to select the questions and 3C1 * 4C1 * 4C1 * 4C1 ways to answer the questions
= 2 * 1 * 3 * 4 * 4 * 4
= 6 * 64
= 384

Total = 384 + 432
= 816

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