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# PS - Combi

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Director
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27 Nov 2006, 09:25
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In a certain test, a student must answer 8 out of 10 questions. How many choices are there if at least 4 of the first five questions must be answered?
Senior Manager
Joined: 01 Sep 2006
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27 Nov 2006, 09:36
In a certain test, a student must answer 8 out of 10 questions. How many choices are there if at least 4 of the first five questions must be answered

4 from 1st and 4 from 2nd = 5C4*5C4=25
4 from 1st and 5 from 2nd = 5c4*5c5=5
5 from 1st and 3 from 2nd = 5C5*5c3=10 coorected for combi
5 from 1st and 4 from 2nd = 5c5*5c4=5
5 from 1st and 5 from 2nd =5c5*5c5=1

[Total 41

Last edited by Damager on 27 Nov 2006, 10:06, edited 1 time in total.
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27 Nov 2006, 09:38
If student has to answer ONLY 8 out of 10 then:

1) Number of ways of answering 4 out of first 5 Qs = 5C4 = 5

2) Number of ways of answering 5 out of first 5 Qs = 5C5 = 1

Number of ways of answering 4 (from 1) out of last 5 Qs = 5C4 = 5

Number of ways of answering 3 (from 2) out of last 5 Qs = 5C3 = 10

Total number of ways if 4 out of first 5 Qs are answered = 5*5 = 25

Total number of ways if 5 out of first 5 Qs are answered = 1*10 = 10

25+10 = 35 total possibilities

OA?
VP
Joined: 15 Jul 2004
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Schools: Wharton (R2 - submitted); HBS (R2 - submitted); IIMA (admitted for 1 year PGPX)
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27 Nov 2006, 11:31
karlfurt wrote:
In a certain test, a student must answer 8 out of 10 questions. How many choices are there if at least 4 of the first five questions must be answered?

If he chooses 4 from the first 5 ques and remaining 4 from last 5

5C4*5C4 = 25

If he chooses 5 from first 5 and 3 from last 5

5C5*5C3 = 1*10 = 10

Total choices the student has are = 25 + 10 = 35
VP
Joined: 15 Jul 2004
Posts: 1473
Schools: Wharton (R2 - submitted); HBS (R2 - submitted); IIMA (admitted for 1 year PGPX)
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27 Nov 2006, 11:35
Damager wrote:
In a certain test, a student must answer 8 out of 10 questions. How many choices are there if at least 4 of the first five questions must be answered

4 from 1st and 4 from 2nd = 5C4*5C4=25
4 from 1st and 5 from 2nd = 5c4*5c5=5
5 from 1st and 3 from 2nd = 5C5*5c3=10 coorected for combi
5 from 1st and 4 from 2nd = 5c5*5c4=5
5 from 1st and 5 from 2nd =5c5*5c5=1

[Total 41

Looks like you misread the question. The number of questions to be answered is only 8.
Director
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28 Nov 2006, 01:42
It comes from an mjj. No OE but 35 seems to be the correct answer. Same as the one provided with the problem.
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