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# Urgent help with math problem

Rinkesh (BBA)     16 March 2012

##### Rinkesh
BBA
41 points

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Below is the question:

A question paper contains 6 questions each having an alternative. The number of ways an examiner can answer one or more questions is:

Kindly help me out on how the question is to be solved. Anubhav (Chartered Accountant)     17 March 2012

##### Anubhav
Chartered Accountant
34 likes  429 points

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By following the given alternative. As there is just one alternative CA Sahil Singla.. (Service Tax )     17 March 2012

##### CA Sahil Singla..
Service Tax
281 likes  3745 points

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Dear Rinkesh

SInce each question has an alternative so there are 3 ways to answer each of the six questions -

1. 1st Alternative

or

2. 2nd Alternative

or

So there are 3 possiblities for each of the 6 questions. Therefore total no. of ways in which a question paper can be answered = 3*3*3*3*3*3 = 729

But in ur case atleast one question must be answered.

The way in which no answer is given to any of the question comes to 1

So in ur case answer = 729-1 =728 Maths Hospital vipin sir   11 October 2018

##### Maths Hospital vipin sir

4 points

View Profile | My Other Post Maths Hospital vipin sir   11 October 2018

##### Maths Hospital vipin sir

4 points

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In how many ways you can answer one or more questions out of 6 questions each having an alternative? This question is useful (1) This question is not useful (0) Comment Answer permutations & combinations 2 years ago, Saheb 1 Answer Solution 1 Each question have 3 choices (answer the question or answer its alternative or leave the question). Therefore, total number of ways = 3 × 3 × 3 × 3 × 3 × 3 = 3 6 But in these 3 6 ways, there is one way in which no question is attended. Therefore, required number of ways = 3 6 − 1 = 728 Solution 2 Select any one question (6C1 ways). Answer the question or alternative (2 ways). Select any two questions (6C2 ways). Each question can be answered in two ways(because either answer the question or its alternative) (total 22 ways). so on ... Therefore, required number of ways = 6C1×2 + 6C2×22 + 6C3×23 + 6C4×24 + 6C5×25 + 6C6×26 = 6 × 2 + 15 × 4 + 20 × 8 + 15 × 16 + 6 × 32 + 1 × 64 = 728 view more »