2 votes 2 votes Consider $30$ multiple-choice questions, each with four options of which exactly one is correct. Then the number of ways one can get only the alternate questions correctly answered is $3^{15}$ $2^{31}$ $2 \times \begin{pmatrix} 30 \\ 15 \end{pmatrix}$ $2 \times 3^{15}$ Combinatory isi2014-dcg combinatory + – Arjun asked Sep 23, 2019 • recategorized Nov 12, 2019 by Lakshman Bhaiya Arjun 975 views answer comment Share Follow See all 13 Comments See all 13 13 Comments reply `JEET commented Oct 1, 2019 reply Follow Share @techbd123 Hey would you like to try this que? 0 votes 0 votes techbd123 commented Oct 2, 2019 reply Follow Share The question seems a bit ambiguous to me by the term "only the alternate questions correctly". What does it mean actually? Does it mean like this below? CA= Correctly Answered IA= Incorrectly Answered NA=Not Answered 1 2 3 4 5 6 ... 28 29 30 CA IA or NA CA IA or NA CA IA or NA ... IA or NA CA IA or NA Here's $2^{15}$ cases. Again 1 2 3 4 5 6 ... 28 29 30 IA or NA CA IA or NA CA IA or NA CA ... CA IA or NA CA Here's $2^{15}$ cases. Total $=2^{15}+2^{15}=2(2^{15})=2^{16}$ 1 votes 1 votes `JEET commented Oct 2, 2019 reply Follow Share Yeah, same doubt I had as well. 1 votes 1 votes Verma Ashish commented Nov 12, 2019 reply Follow Share Two ways for answering all the questions such that alternate answers are correct-- $T F T F \ldots TF$ $FTFT\ldots FT$ ==> $3^{15}+3^{15}$ 3 votes 3 votes `JEET commented Nov 12, 2019 reply Follow Share Why $3$ as the base? @Verma Ashish 1 votes 1 votes Verma Ashish commented Nov 12, 2019 reply Follow Share There are 3 ways to choose wrong option, for each question. And only 1 way to choose correct option Because only one choice is correct out of four options. 0 votes 0 votes `JEET commented Nov 12, 2019 i edited by `JEET Nov 12, 2019 reply Follow Share Then how would you explain this statement? Then the number of ways one can get only the alternate questions correctly answered is The question asks for true answers, right? You considered for false only? 1 votes 1 votes imShreyas commented Nov 12, 2019 reply Follow Share There is only one correct option for right answer, whereas there were 3 different ways to answer it wrong. Since you had to get alternate questions right, you could start with either a right answer or wrong. Can you please elaborate your doubt? 1 votes 1 votes chirudeepnamini commented Nov 21, 2019 reply Follow Share @imShreyas I thought in a same way but i have doubt.. This is the way i solved the problem: let 1,2,3,4...30 be the questions and each question has options a,b,c,d. to satisfy the question student must answer,in one of following ways. 1,3,5,7,9,...29 or 2,4,6,8,...,30 why only these 2 sequences? Let us start with 3 and answering alternate questions correctly.. then we have sequence 3,5,7,...29 but we have solved only 14 questions so he has to choose one number from 1 or 2 but only 1 satisfy the condition. So the sequence formed is again 1,3,5,...29 Let us start with 9 and answering alternate questions correctly.. then we have sequence 9,11,13,15...29 but we have solved only 11 questions so he has to choose 4 numbers from 1 to 8 but only 1,3,5,7 satisfy the condition. So the sequence formed is again 1,3,5,...29 similarly we can prove for 2,4,6,...30. So even if we try to form a different sequence, we end up only in one of these 2 sequences.. and Now consider the sequence 1,3,5,....29 each of this question has to be answered correctly and all other questions has to be wrong. so each of 1,3,5...29 is to be answered in 1 way and each of 2,4,6...30 has 3 choices to be answered.. so we have $3^{15}$ ways. similarly we have $3^{15}$ ways for other sequence.. So on total there are $3^{15}+3^{15}$ ways.. which is option c but my doubt is why are we not considering the order in which questions are answered. I mean each sequence can be answered in $15!$ ways.. so why can't the answer be $2*3^{15}*15!$ where am i going wrong? 0 votes 0 votes Shaktiman commented Jan 4, 2020 reply Follow Share Let solve this question to divide into smaller part. Let only we have 4 Question . 1) a,b,c,d(4 option ,a is correct here ) 2)a,b,c,d(a is correct here ) 3) a,b,c,d(b is correct here ) 4)a,b,c,d(a is correct here ) So the alternate answer should be right . It can be possible in two ways . Case 1 1) false 2)true 3)false 4)true In true case i have only one option but in flase option we have 3 option total possible way. 3^2. Case 2) 1)true 2) false 3)true 4)false Same as case 1. 3^2. Total possible way. 2*3^2 (for four questions) For 30 question it should be 2*3^15. 0 votes 0 votes Hradesh patel commented Jan 4, 2020 reply Follow Share What are three option of false.??? I think its ( wrong, not attempted) 2 ways Plz explain?? 0 votes 0 votes Verma Ashish commented Jan 4, 2020 reply Follow Share Out of 4 options for each question only one option is correct..rest 3 are false 1 votes 1 votes Hradesh patel commented Jan 4, 2020 reply Follow Share Ok,...i missed.. thanks 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes The answer pattern can be: CWCWCW…...CW, or WCWCWC…...WC For first case: each W has 3 choices and each C has only 1 choice => 3^15 choices for second case: 3^15 case, from similar logic as first case. so, total ways = 2*(3^15) neeraj_bhatt answered Sep 5, 2020 neeraj_bhatt comment Share Follow See all 0 reply Please log in or register to add a comment.