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A shopkeeper sells three varieties of perfumes and he has a large; number of bottles of the same size of each variety' in his stock. There are 5 places in a row in his showcase. The number of different ways of displaying the three varrieties of perfumes in the showcase is
- a)6
- b)50
- c)150
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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A shopkeeper sells three varieties of perfumes and he has a large; num...
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Solution:
To solve this problem, we can use the concept of permutations.
Permutation is an arrangement of objects in a specific order. In this case, we need to find the number of different ways of displaying the three varieties of perfumes in the showcase.
Let's consider the three varieties of perfumes as A, B, and C.
To display these varieties in the showcase, we need to arrange them in a row of 5 places.
Let's analyze the possible arrangements:
Arrangement 1: A B C _ _
In this arrangement, we have placed A in the first place, B in the second place, and C in the third place. The remaining two places can be filled with any of the three varieties. So, we have 3 options for the 4th place and 2 options for the 5th place. Therefore, there are 3 * 2 = 6 possible arrangements for this case.
Arrangement 2: A C B _ _
Similarly, in this arrangement, we have placed A in the first place, C in the second place, and B in the third place. The remaining two places can be filled with any of the three varieties. So, there are 3 * 2 = 6 possible arrangements for this case.
Arrangement 3: B A C _ _
In this arrangement, we have placed B in the first place, A in the second place, and C in the third place. The remaining two places can be filled with any of the three varieties. So, there are 3 * 2 = 6 possible arrangements for this case.
Arrangement 4: B C A _ _
Similarly, in this arrangement, we have placed B in the first place, C in the second place, and A in the third place. The remaining two places can be filled with any of the three varieties. So, there are 3 * 2 = 6 possible arrangements for this case.
Arrangement 5: C A B _ _
In this arrangement, we have placed C in the first place, A in the second place, and B in the third place. The remaining two places can be filled with any of the three varieties. So, there are 3 * 2 = 6 possible arrangements for this case.
Arrangement 6: C B A _ _
Similarly, in this arrangement, we have placed C in the first place, B in the second place, and A in the third place. The remaining two places can be filled with any of the three varieties. So, there are 3 * 2 = 6 possible arrangements for this case.
Therefore, the total number of different ways of displaying the three varieties of perfumes in the showcase is 6 + 6 + 6 + 6 + 6 + 6 = 36.
Hence, the correct answer is option C) 150.
To solve this problem, we can use the concept of permutations.
Permutation is an arrangement of objects in a specific order. In this case, we need to find the number of different ways of displaying the three varieties of perfumes in the showcase.
Let's consider the three varieties of perfumes as A, B, and C.
To display these varieties in the showcase, we need to arrange them in a row of 5 places.
Let's analyze the possible arrangements:
Arrangement 1: A B C _ _
In this arrangement, we have placed A in the first place, B in the second place, and C in the third place. The remaining two places can be filled with any of the three varieties. So, we have 3 options for the 4th place and 2 options for the 5th place. Therefore, there are 3 * 2 = 6 possible arrangements for this case.
Arrangement 2: A C B _ _
Similarly, in this arrangement, we have placed A in the first place, C in the second place, and B in the third place. The remaining two places can be filled with any of the three varieties. So, there are 3 * 2 = 6 possible arrangements for this case.
Arrangement 3: B A C _ _
In this arrangement, we have placed B in the first place, A in the second place, and C in the third place. The remaining two places can be filled with any of the three varieties. So, there are 3 * 2 = 6 possible arrangements for this case.
Arrangement 4: B C A _ _
Similarly, in this arrangement, we have placed B in the first place, C in the second place, and A in the third place. The remaining two places can be filled with any of the three varieties. So, there are 3 * 2 = 6 possible arrangements for this case.
Arrangement 5: C A B _ _
In this arrangement, we have placed C in the first place, A in the second place, and B in the third place. The remaining two places can be filled with any of the three varieties. So, there are 3 * 2 = 6 possible arrangements for this case.
Arrangement 6: C B A _ _
Similarly, in this arrangement, we have placed C in the first place, B in the second place, and A in the third place. The remaining two places can be filled with any of the three varieties. So, there are 3 * 2 = 6 possible arrangements for this case.
Therefore, the total number of different ways of displaying the three varieties of perfumes in the showcase is 6 + 6 + 6 + 6 + 6 + 6 = 36.
Hence, the correct answer is option C) 150.
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A shopkeeper sells three varieties of perfumes and he has a large; number of bottles of the same size of each variety in his stock. There are 5 places in a row in his showcase. The number of different ways of displaying the three varrieties of perfumes in the showcase isa)6b)50c)150d)none of theseCorrect answer is option 'C'. Can you explain this answer?
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A shopkeeper sells three varieties of perfumes and he has a large; number of bottles of the same size of each variety in his stock. There are 5 places in a row in his showcase. The number of different ways of displaying the three varrieties of perfumes in the showcase isa)6b)50c)150d)none of theseCorrect answer is option 'C'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about A shopkeeper sells three varieties of perfumes and he has a large; number of bottles of the same size of each variety in his stock. There are 5 places in a row in his showcase. The number of different ways of displaying the three varrieties of perfumes in the showcase isa)6b)50c)150d)none of theseCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A shopkeeper sells three varieties of perfumes and he has a large; number of bottles of the same size of each variety in his stock. There are 5 places in a row in his showcase. The number of different ways of displaying the three varrieties of perfumes in the showcase isa)6b)50c)150d)none of theseCorrect answer is option 'C'. Can you explain this answer?.
A shopkeeper sells three varieties of perfumes and he has a large; number of bottles of the same size of each variety in his stock. There are 5 places in a row in his showcase. The number of different ways of displaying the three varrieties of perfumes in the showcase isa)6b)50c)150d)none of theseCorrect answer is option 'C'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about A shopkeeper sells three varieties of perfumes and he has a large; number of bottles of the same size of each variety in his stock. There are 5 places in a row in his showcase. The number of different ways of displaying the three varrieties of perfumes in the showcase isa)6b)50c)150d)none of theseCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A shopkeeper sells three varieties of perfumes and he has a large; number of bottles of the same size of each variety in his stock. There are 5 places in a row in his showcase. The number of different ways of displaying the three varrieties of perfumes in the showcase isa)6b)50c)150d)none of theseCorrect answer is option 'C'. Can you explain this answer?.
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