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17 Mar 2021, 05:00
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B
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:
27% (02:45) correct
73%
(03:12)
wrong
based on 71
sessions
History
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Time
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Not Attempted Yet
Amelia, Betty, Cindy, Dorothy, Elly and Frida are seated on chairs numbered 1 to 6 respectively. Their birthdays fall on the \(21^{st}\), \(2^{nd}\), \(19^{th}\), \(31^{st}\), \(18^{th}\) and \(5^{th}\), respectively of the same month. If the seat number of a lady matches with the remainder of her birth date when the birth date is divided by 15, we say that she is “properly seated” else we say that she is “wrongly seated”. In how many other ways could these wrongly seated ladies be ‘wrongly seated’?
(A) 9
(B) 43
(C) 44
(D) 45
(E) 74
(A) 9
(B) 43
(C) 44
(D) 45
(E) 74
18 Mar 2021, 00:01
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Given: A, B, C, D, E and F are seated on chairs 1 to 6 respectively (taking initials of name only). Their birthdays fall on 21, 2, 19, 31, 18 and 5 respectively.
Remainders when birthdays are divided by 15 are 6, 2, 4, 1, 3 and 5 respectively.
Seat number of B only matches with remainder i.e.2, so B is properly seated, all others are wrongly seated.
=> We have 5 people and 5 chairs. We will not consider B here.
Other ways in which wrongly seated ladies could be wrongly seated =?
Let us go step by step now:
If we have 2 people and 2 chairs, then they can be wrongly seated in 1 way only.
If we have 3 people and 3 chairs, then they all can be wrongly seated in 2 ways.
For eg: Properly seated - ABC
Possibilities: ABC, ACB, BCA, BAC, CAB, CBA
So, only ABC is properly seated, but ACB, BAC and CBA are possibilities in which at least one is properly seated.
We want all to be wrongly seated - BCA and CAB - 2 possibilities.
=> 3 people can all be wrongly seated in 2 ways
If we have 4 people and 4 chairs, then above process of writing possibilities becomes tedious.
So we can do one thing.
Possibilities of all wrongly seated + Possibilities of at least one properly seated = Total possibilities
Total possibilities of arranging 4 people = 4! = 24
Possibilities of at least one properly seated = 4C4 + 4C2 * 1 + 4C1 * 2 = 1 + 6 * 1 + 4 * 2 = 15 (4C3 is not taken because if 3 are properly seated, then 4th will automatically be properly seated, i.e. that case is already counted)
Possibilities of all wrongly seated = Total possibilities - Possibilities of at least one properly seated = 24 - 15 = 9
=> 4 people can all be wrongly seated in 9 ways
Now let us jump to our question.
Possibilities of all wrongly seated + Possibilities of at least one properly seated = Total possibilities
Total possibilities of arranging 5 people = 5! = 120
Possibilities of at least one properly seated = 5C5 + 5C3 * 1 + 5C2 * 2 + 5C1 * 9 = 1 + 10 * 1 + 10 * 2 + 5 * 9 = 76 (5C4 is not taken because if 4 are properly seated, then 5th will automatically be properly seated, i.e. that case is already counted)
Possibilities of all wrongly seated = Total possibilities - Possibilities of at least one properly seated = 120 - 76 = 44
=> 5 people can all be wrongly seated in 44 ways.
Out of these 44 ways, 1 arrangement is given i.e. 6, 2, 4, 1, 3, 5.
Remaining arrangements = 44 - 1 = 43 ways.
Therefore, wrongly seated ladies could be wrongly seated in other 43 ways.
So, correct answer is option B.
Remainders when birthdays are divided by 15 are 6, 2, 4, 1, 3 and 5 respectively.
Seat number of B only matches with remainder i.e.2, so B is properly seated, all others are wrongly seated.
=> We have 5 people and 5 chairs. We will not consider B here.
Other ways in which wrongly seated ladies could be wrongly seated =?
Let us go step by step now:
If we have 2 people and 2 chairs, then they can be wrongly seated in 1 way only.
If we have 3 people and 3 chairs, then they all can be wrongly seated in 2 ways.
For eg: Properly seated - ABC
Possibilities: ABC, ACB, BCA, BAC, CAB, CBA
So, only ABC is properly seated, but ACB, BAC and CBA are possibilities in which at least one is properly seated.
We want all to be wrongly seated - BCA and CAB - 2 possibilities.
=> 3 people can all be wrongly seated in 2 ways
If we have 4 people and 4 chairs, then above process of writing possibilities becomes tedious.
So we can do one thing.
Possibilities of all wrongly seated + Possibilities of at least one properly seated = Total possibilities
Total possibilities of arranging 4 people = 4! = 24
Possibilities of at least one properly seated = 4C4 + 4C2 * 1 + 4C1 * 2 = 1 + 6 * 1 + 4 * 2 = 15 (4C3 is not taken because if 3 are properly seated, then 4th will automatically be properly seated, i.e. that case is already counted)
Possibilities of all wrongly seated = Total possibilities - Possibilities of at least one properly seated = 24 - 15 = 9
=> 4 people can all be wrongly seated in 9 ways
Now let us jump to our question.
Possibilities of all wrongly seated + Possibilities of at least one properly seated = Total possibilities
Total possibilities of arranging 5 people = 5! = 120
Possibilities of at least one properly seated = 5C5 + 5C3 * 1 + 5C2 * 2 + 5C1 * 9 = 1 + 10 * 1 + 10 * 2 + 5 * 9 = 76 (5C4 is not taken because if 4 are properly seated, then 5th will automatically be properly seated, i.e. that case is already counted)
Possibilities of all wrongly seated = Total possibilities - Possibilities of at least one properly seated = 120 - 76 = 44
=> 5 people can all be wrongly seated in 44 ways.
Out of these 44 ways, 1 arrangement is given i.e. 6, 2, 4, 1, 3, 5.
Remaining arrangements = 44 - 1 = 43 ways.
Therefore, wrongly seated ladies could be wrongly seated in other 43 ways.
So, correct answer is option B.
03 Nov 2024, 20:20
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In the 4 people, 4 chairs scenario, why are you multiplying by 1 and 2 in this equation:
Possibilities of at least one properly seated = 4C4 + 4C2 * 1 + 4C1 * 2 = 1 + 6 * 1 + 4 * 2 = 15
Thanks!
Possibilities of at least one properly seated = 4C4 + 4C2 * 1 + 4C1 * 2 = 1 + 6 * 1 + 4 * 2 = 15
Thanks!
MarmikUpadhyay
