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by parveen110 » Sun Jan 19, 2014 4:57 am
The number of all possible selections which a student can make for answering one or more questions out of 10 given questions in a paper, when each question has an alternative is:
a. 1345
b. 23560
c. 541340
d. 59048
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by [email protected] » Sun Jan 19, 2014 4:15 pm
Hi parveen110,

This is NOT a GMAT question and also happens to be poorly worded. It is unclear what the phrase "each question has an alternative" means. If the question asked for "the number of combinations of questions that could be answered if at last one question was answered", then this would be a multi-step combinatorics question (and would require several small calculations).

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by parveen110 » Sun Jan 19, 2014 8:32 pm
[email protected] wrote:Hi parveen110,

This is NOT a GMAT question and also happens to be poorly worded. It is unclear what the phrase "each question has an alternative" means. If the question asked for "the number of combinations of questions that could be answered if at last one question was answered", then this would be a multi-step combinatorics question (and would require several small calculations).

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Hi Rich,

Thanks for the reply.However, This question doesn't involve multi-step combinatorics. And the phrase "each question has an alternative" means that each question has two choices and out of which only one can be chosen.

I have a solution which i'm little confused about. Here it goes:

Each question can be attempted in three ways:
1. The student do not solve the question.
2. The student attempts first part of the question.
3. The student attempts the second(i.e.,alternative) part of the question.

Therefore required number of ways= 3^10-1=59048.

My confusion regarding the solution part is:

Can i not solve it in a following way:

The student may select 1 question out of 10 in 10C1 ways.

The student may select 2 question out of 10 in 10C2 ways.

and so on...
Finally the student may select 10 questions out of 10 in 10C10 ways.

Adding all the possibilities up:

(10C0+10C1+10C2+10C3+10C4+....+10C10) = 2^10 = 1024

Now since each question has an alternative, I multiply the above step by 2= 2 x 1024
=4028 ways.
Now since according to the question, i have to attempt at least 1 question,
I may do so in 4028-1 = 4027 ways.

I just would like to know where is this approach flawed. Am i making not accounting for something by multiplying the possibilities by 2?

Hope I've made myself amply clear:)

Thanks.
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