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03 May 2025, 00:55
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
52% (02:44) correct
48%
(02:50)
wrong
based on 165
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In an entrance exam, there are 3 sections viz, Math, English, and Social. Each section has 3 questions. A student needs to attempt a total of 5 questions such that he/she attempts at least one question from each section. In how many ways it can be done?
A. 27
B. 54
C. 81
D. 108
E. 162
A. 27
B. 54
C. 81
D. 108
E. 162
03 May 2025, 04:35
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Bhagi123
Total no. of ways to select 5 questions out of 9 = 9C5 = 18*7
No. of possibilities omitting 1 section:
Math and English: 3 Q's from Math = 1 way and 2 Q's from English = 3 ways; similarly, if it were 2 Q's from Math = 3 ways and 3 Q's from English = 1 way. So in total if we omit Social there are 6 possible ways to attempt questions
As we did above, if we omit English = 6 possible ways and if we omit Math = 6 possible ways.
Therefore, no. of possibilities omitting 1 section: 6 + 6 + 6 = 18 ways
Ways of attempting atleast one question from each section = Total no. of ways to select 5 questions out of 9 - No. of possibilities omitting 1 section
= 18*7 - 18
= 18*6
= 108
Answer D.
08 Sep 2025, 10:06
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Two ways:
a) 2 + 2 + 1
Select the two subjects from which 2 questions are to be attempted = 3C2
For each of these 2 subjects 3C2 choices of selecting 2 questions =3C2 * 3C2
For 3rd subject, 3C1 choices to select the 1 question = 3C1
Total choices = 3C2 * 3C2 * 3C2 * 3C1 = 81
b) 3 + 1 + 1
Select the one subject from which 3 questions will be attempted = 3C1
Choice of questions = 3C3.
For the remaining 2 subjects, choices = 3C1 * 3C1
Total choices = 3C1 * 1 * 3C1 * 3C1 = 27
Answer = 81 + 27 = 108
a) 2 + 2 + 1
Select the two subjects from which 2 questions are to be attempted = 3C2
For each of these 2 subjects 3C2 choices of selecting 2 questions =3C2 * 3C2
For 3rd subject, 3C1 choices to select the 1 question = 3C1
Total choices = 3C2 * 3C2 * 3C2 * 3C1 = 81
b) 3 + 1 + 1
Select the one subject from which 3 questions will be attempted = 3C1
Choice of questions = 3C3.
For the remaining 2 subjects, choices = 3C1 * 3C1
Total choices = 3C1 * 1 * 3C1 * 3C1 = 27
Answer = 81 + 27 = 108
