Computer-Networks
October 4, 2023Operating-Systems
October 4, 2023UGC NET CS 2016 July- paper-2
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Question 1
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How many different equivalence relations with exactly three different equivalence classes are there on a set with five elements?
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10
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15
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25
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30
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Question 1 Explanation:
Step-1: Given number of equivalence classes with 5 elements with three elements in each class will be 1,2,2 (or) 2,1,2 (or) 2,2,1 and 3,1,1.
Step-2: The number of combinations for three equivalence classes are
2,2,1 chosen in ( 5 C 2 * 3 C 2 * 1 C 1 )/2! = 15
3,1,1 chosen in( 5 C 2 * 3 C 2 * 1 C 1 )/2! = 10
Step-3: Total differential classes are 15+10
=25.
Step-2: The number of combinations for three equivalence classes are
2,2,1 chosen in ( 5 C 2 * 3 C 2 * 1 C 1 )/2! = 15
3,1,1 chosen in( 5 C 2 * 3 C 2 * 1 C 1 )/2! = 10
Step-3: Total differential classes are 15+10
=25.
Correct Answer: C
Question 1 Explanation:
Step-1: Given number of equivalence classes with 5 elements with three elements in each class will be 1,2,2 (or) 2,1,2 (or) 2,2,1 and 3,1,1.
Step-2: The number of combinations for three equivalence classes are
2,2,1 chosen in ( 5 C 2 * 3 C 2 * 1 C 1 )/2! = 15
3,1,1 chosen in( 5 C 2 * 3 C 2 * 1 C 1 )/2! = 10
Step-3: Total differential classes are 15+10
=25.
Step-2: The number of combinations for three equivalence classes are
2,2,1 chosen in ( 5 C 2 * 3 C 2 * 1 C 1 )/2! = 15
3,1,1 chosen in( 5 C 2 * 3 C 2 * 1 C 1 )/2! = 10
Step-3: Total differential classes are 15+10
=25.