plz slv it
An examination paper is divided into three sections. Each section contain six questions, three of which are starred. The rubric says “Answer only one question from each section. Answer at least one starred question and at least one unstarred question”. How many different choices are possible?
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An examination paper is divided into three sections. Each section contain six questions, three of which are starred. The rubric says “Answer only one question from each section. Answer at least one starred question and at least one unstarred question”. How many different choices are possible?
Possible Combination:
1 starred and 2 non-starred
2 starred and 1 non-starred
each case select in 3c1 x 3c1 x 2c1 x 3c1 x 1c1 x 3c1=162
(first select any one section out of three(3c1) then, from selected section select one question out of 3(3c1), then select another section out of 2(remain 2 section)(2c1), then from selected second section select one question out of three(3c1), now only one section remain(1c1), now from third section select one question out of three(3c1)
so total ways =162+162=324
correct me if i m wrong
No one remain as a vergin because life fucks everyone
Little Star
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I guess secondharsa is correct !!!
first select any one section out of three(3c1) then, from selected section select one question out of 3(3c1), then select another section out of 2(remain 2 section)(2c1), then from selected second section select one question out of three(3c1), now only one section remain(1c1), now from third section select one question out of three(3c1)
While selecting the questions you are not considering if they are starred or not and thus multiplying an extra 3
It should be 54 not 162
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if at least one starred and one unstarred question is the required case. we will take its complement that is if all are starred or all are unstarred.
for this complementary case no. of options are 3x3x3=27
total no. of cases with one question from each section will be 6x6x6=216
therefore our requisite ans will be 216-27=189
Possible ways of attempting the questions
1 starred + 2 non-starred
2 starred + 1 non-starred
But Number of ways of attempting the questions in both the case should be same as the number of starred and non-starred questions are same
Number of ways of selecting a section for starred questions = 3 and after that we can chose one question out of the available 3 in 3 diff. ways
So possible comb = 3 x 3
Similarly number of ways of selecting the non-starred questions = 2 x 3
Total possible comb for each case = 9 x 6 = 54
So number of ways of attempting the questions = 2 x 54 = 108
But I am not sure abt the answer