Ooh, this is a great problem

So, let's call the questions A, B, C, D, and E, and let's assign the number of answer choices for each as follows: A(3), B(3), C(4), D(4), E(4).

You get to omit one question out of the five, so all of the following choices are possible as far as questions to be done:

A(3), B(3), C(4), D(4), E(4).
A(3), B(3), C(4), D(4), E(4).
A(3), B(3), C(4), D(4), E(4).
A(3), B(3), C(4), D(4), E(4).
A(3), B(3), C(4), D(4), E(4).

For of these individual scenarios, we can calculate the number of possible overall answer combinations through simple multiplication (since each question is independent of each other one). So we'll have:

A(3), B(3), C(4), D(4), E(4). 3*4*4*4 = 192
A(3), B(3), C(4), D(4), E(4). 3*4*4*4 = 192
A(3), B(3), C(4), D(4), E(4). 3*3*4*4 = 144
A(3), B(3), C(4), D(4), E(4). 3*3*4*4 = 144
A(3), B(3), C(4), D(4), E(4). 3*3*4*4 = 144.

Then, since any of these five scenarios (of which question to omit) is possible, we add those numbers together. 2*192 + 3*144 = 816.

And there you go! Does that make sense?

SwatiDenre wrote:A teacher prepares a test. She gives 5 objective types questions out of which 4 are 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



Ans - 816

How to solve this.
Thanks in Advance.

here two cases are possible;
case 1) when students answer first two question and remaining two question from the last three question.

case 2) when students answer one from first two question and remaining three from the last three questions.

case 1)first two question can be selected in 2C2 ways, and since each question has 3 options therefore total no. of ways of answering first two=2C2*3C1*3C1 also we can select any two question from the last three question in 3C2 and as each of the last three question has 4 answer choices therefore total no. of ways of answering last two=3C2*4C1*4C1;
hence total no. of ways of case 1= 2C2*3C1*3C1*3C2*4C1*4C1=432

case 2) one question from first two can be selected and answered in 2C1*3C1; and 3 question from last three can be selected and answered in 3C3*4C1*4C1*4C1;
hence total no. of ways of case 2= 2C1*3C1*3C3*4C1*4C1*4C1=384

hence total no. of ways= case1)+ case 2)= 432+384=816 hence B