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19 Aug 2022, 09:09
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D
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Difficulty:
95%
(hard)
Question Stats:
28% (01:51) correct
72%
(02:03)
wrong
based on 148
sessions
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How many arch-enemies does Ronaldo have ?
(1) Ronaldo could make more than 469 but fewer than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).
(2) Ronaldo could make more than 511 but fewer than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).
(1) Ronaldo could make more than 469 but fewer than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).
(2) Ronaldo could make more than 511 but fewer than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).
19 Aug 2022, 09:09
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Official Solution:
How many arch-enemies does Ronaldo have ?
(1) Ronaldo could make more than 469 but fewer than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).
Say Ronaldo has \(n\) arch-enemies: \(\{a_1, \ a_2, \ a_3, \ ...a_n\}\). Each of \((a_1, \ a_2, \ a_3, \ ...a_n)\) has TWO options: either to be included in the list or not. Hence the total number of lists Ronaldo could make is \(2^n\).
But notice that, this number will include lists with 1 enemy, lists with 2 enemies and an empty list with 0 enemies in it. Since the lists should contain at least 2 subordinates, then we should subtract all the lists containing only 1 enemy, lists with 2 enemies and an empty list from \(2^n\).
The number of lists with 1 enemy is \(n\): \(\{a_1, \ 0, \ 0, \ 0, \ ..., \ 0\}\), \(\{0, \ a_2, \ 0, \ 0, \ ..., \ 0\}\), \(\{0, \ 0, \ a_3, \ 0, \ ..., \ 0\}\), ..., \(\{0, \ 0, \ 0, \ ..., \ a_n\}\).
The number of lists with 2 enemies is \(C^2_n = \frac{n!}{2!(n-2)!}=\frac{(n-1)n}{2}\).
The number of lists with 0 enemies (empty list) is 1: \(\{0, \ 0, \ 0, \ ..., \ 0\}\)
So, we have that the number of lists with at least 3 arch-enemies is \(2^n - n - \frac{(n-1)n}{2}-1\)
We are told that \(469 < 2^n - n - \frac{(n-1)n}{2}-1 < 1069 \).
Add 1 to all three parts: \(470 < 2^n - n - \frac{(n-1)n}{2} < 1070 \).
Now, test some numbers. We know that \(2^{9}=512\) and \(2^{10}=1024\), so \(n\) should be somewhere in that neighborhood:
If \(n = 9\), then \(2^n - n - \frac{(n-1)n}{2}=512-9-36=467\). Not good!
If \(n = 10\), then \(2^n - n - \frac{(n-1)n}{2}=1024-10-45=969\) Good!
If \(n = 11\), then \(2^n - n - \frac{(n-1)n}{2}=2048-11-55=1982\). Not good!
Sufficient.
(2) Ronaldo could make more than 511 but fewer than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).
Apply the same logic as in (1), the difference being that now we are interested in the lists containing at least 2 enemies instead of the lists containing at least 3 enemies:
\(511 < 2^n - n - 1 < 1025\)
By testing the values, we can get that the above holds true only for \(n=10\).
Sufficient.
Answer: D
How many arch-enemies does Ronaldo have ?
(1) Ronaldo could make more than 469 but fewer than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).
Say Ronaldo has \(n\) arch-enemies: \(\{a_1, \ a_2, \ a_3, \ ...a_n\}\). Each of \((a_1, \ a_2, \ a_3, \ ...a_n)\) has TWO options: either to be included in the list or not. Hence the total number of lists Ronaldo could make is \(2^n\).
But notice that, this number will include lists with 1 enemy, lists with 2 enemies and an empty list with 0 enemies in it. Since the lists should contain at least 2 subordinates, then we should subtract all the lists containing only 1 enemy, lists with 2 enemies and an empty list from \(2^n\).
The number of lists with 1 enemy is \(n\): \(\{a_1, \ 0, \ 0, \ 0, \ ..., \ 0\}\), \(\{0, \ a_2, \ 0, \ 0, \ ..., \ 0\}\), \(\{0, \ 0, \ a_3, \ 0, \ ..., \ 0\}\), ..., \(\{0, \ 0, \ 0, \ ..., \ a_n\}\).
The number of lists with 2 enemies is \(C^2_n = \frac{n!}{2!(n-2)!}=\frac{(n-1)n}{2}\).
The number of lists with 0 enemies (empty list) is 1: \(\{0, \ 0, \ 0, \ ..., \ 0\}\)
So, we have that the number of lists with at least 3 arch-enemies is \(2^n - n - \frac{(n-1)n}{2}-1\)
We are told that \(469 < 2^n - n - \frac{(n-1)n}{2}-1 < 1069 \).
Add 1 to all three parts: \(470 < 2^n - n - \frac{(n-1)n}{2} < 1070 \).
Now, test some numbers. We know that \(2^{9}=512\) and \(2^{10}=1024\), so \(n\) should be somewhere in that neighborhood:
If \(n = 9\), then \(2^n - n - \frac{(n-1)n}{2}=512-9-36=467\). Not good!
If \(n = 10\), then \(2^n - n - \frac{(n-1)n}{2}=1024-10-45=969\) Good!
If \(n = 11\), then \(2^n - n - \frac{(n-1)n}{2}=2048-11-55=1982\). Not good!
Sufficient.
(2) Ronaldo could make more than 511 but fewer than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).
Apply the same logic as in (1), the difference being that now we are interested in the lists containing at least 2 enemies instead of the lists containing at least 3 enemies:
\(511 < 2^n - n - 1 < 1025\)
By testing the values, we can get that the above holds true only for \(n=10\).
Sufficient.
Answer: D
General Discussion
12 Sep 2022, 09:29
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Love this question! Please link similar ones testing hard combinatorics where we consider different subset sizes
