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A student council is to be chosen from a class of 12 student
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Updated on: 15 Dec 2017, 01:18
Updated on: 15 Dec 2017, 01:18
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Question Stats:
54% (01:14) correct
45% (01:35) wrong based on 53 sessions
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A student council is to be chosen from a class of 12 students consisting of a president, a vice president, and 3 committee members. How many such councils are possible
(A) \(\frac{12!}{7!5!}\)
(B) \(\frac{12!}{7!3!}\)
(C) \(\frac{12!}{3!5!}\)
(D) \(\frac{12!}{7!}\)
(E) \(12!\)
(A) \(\frac{12!}{7!5!}\)
(B) \(\frac{12!}{7!3!}\)
(C) \(\frac{12!}{3!5!}\)
(D) \(\frac{12!}{7!}\)
(E) \(12!\)
Re: Counting
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13 Sep 2017, 14:21
13 Sep 2017, 14:21
4
Expert Reply
Hi,
In total there are 5 members to be selected out of 12.
If all 12 members were the same then the choice would have been 12C5, i.e. Option A.
But all members are not the same since there are two maned positions namely president and vice president.
If all the positions were distict the choice 12P5 would be correct, i.e. Option D would have been right.
There are only 2 named positions and 3 non named positions.
No of ways to select a president= 12.
No of ways to select a vice president =11.
No of ways to select 3 members from the remaining 10 is 10C3= \(\frac{10!}{7! \times 3!}\)
Hence number of ways to selct president, vice president and 3 members = \(12 \times 11 \times\frac{10!}{7! \times 3!}\)= \(\frac{12!}{7! \times 3!}\).
Hence option B is correct!
In total there are 5 members to be selected out of 12.
If all 12 members were the same then the choice would have been 12C5, i.e. Option A.
But all members are not the same since there are two maned positions namely president and vice president.
If all the positions were distict the choice 12P5 would be correct, i.e. Option D would have been right.
There are only 2 named positions and 3 non named positions.
No of ways to select a president= 12.
No of ways to select a vice president =11.
No of ways to select 3 members from the remaining 10 is 10C3= \(\frac{10!}{7! \times 3!}\)
Hence number of ways to selct president, vice president and 3 members = \(12 \times 11 \times\frac{10!}{7! \times 3!}\)= \(\frac{12!}{7! \times 3!}\).
Hence option B is correct!
Re: Counting
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10 Jan 2019, 22:35
10 Jan 2019, 22:35
1
Expert Reply
sandy wrote:
Hi,
In total there are 5 members to be selected out of 12.
If all 12 members were the same then the choice would have been 12C5, i.e. Option A.
But all members are not the same since there are two maned positions namely president and vice president.
If all the positions were distict the choice 12P5 would be correct, i.e. Option D would have been right.
There are only 2 named positions and 3 non named positions.
No of ways to select a president= 12.
No of ways to select a vice president =11.
No of ways to select 3 members from the remaining 10 is 10C3= \(\frac{10!}{7! \times 3!}\)
Hence number of ways to selct president, vice president and 3 members = \(12 \times 11 \times\frac{10!}{7! \times 3!}\)= \(\frac{12!}{7! \times 3!}\).
Hence option B is correct!
In total there are 5 members to be selected out of 12.
If all 12 members were the same then the choice would have been 12C5, i.e. Option A.
But all members are not the same since there are two maned positions namely president and vice president.
If all the positions were distict the choice 12P5 would be correct, i.e. Option D would have been right.
There are only 2 named positions and 3 non named positions.
No of ways to select a president= 12.
No of ways to select a vice president =11.
No of ways to select 3 members from the remaining 10 is 10C3= \(\frac{10!}{7! \times 3!}\)
Hence number of ways to selct president, vice president and 3 members = \(12 \times 11 \times\frac{10!}{7! \times 3!}\)= \(\frac{12!}{7! \times 3!}\).
Hence option B is correct!
This was an awesome breakdown, thanks!
