A) 10
B) 9
C) 12
D) 15
Correct Answer: B
Solution :
Pencils \[({{P}_{1}},{{P}_{2}},{{P}_{3}})\] and Erasers \[({{E}_{1}},{{E}_{2}},{{E}_{3}})\] Possible combinations \[({{P}_{1}},{{E}_{1}}),({{P}_{1}},{{E}_{2}}),({{P}_{1}}{{E}_{3}}),({{P}_{2}},{{E}_{1}}),({{P}_{2}},{{E}_{2}}),({{P}_{2}}{{E}_{3}}),({{P}_{3}},{{E}_{1}}),({{P}_{3}},{{E}_{2}}),({{P}_{3}},{{E}_{3}})\] or Three elements are in Pencil set and three elements are in eraser set. So, Possible combinations \[=3\times 3=9.\]You need to login to perform this action.
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