Ben and Ann are among 7 contestants from which 4 : PS Archive
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Ben and Ann are among 7 contestants from which 4

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Ben and Ann are among 7 contestants from which 4 [#permalink]

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17 Sep 2003, 07:57
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Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, How many contain neither Ben nor Ann.

why is 7C4 - 5C2 not correct?
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17 Sep 2003, 09:42
If i'm reading it right the question says how many combinations without ben and Ann. So the answer would be 5C4
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17 Sep 2003, 15:45
rich28 wrote:
If i'm reading it right the question says how many combinations without ben and Ann. So the answer would be 5C4

I agree with Rich. If I understood the question correctly, then 5C4 must be the answer.
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17 Sep 2003, 21:06
edealfan wrote:
rich28 wrote:
If i'm reading it right the question says how many combinations without ben and Ann. So the answer would be 5C4

I agree with Rich. If I understood the question correctly, then 5C4 must be the answer.

yeah, 5is correct

but can you tell me what i did wrong here..

i thought something like this

Total # of ways = 7C4

Consider cases where ben and ann are always present...so we have only 2 people to select from the remaining five

Thats how I got 5C2

so the # of cases where both are not present = 7C4 - 5C2

i cant spot the mistake...please clarify

thanks
praetorian
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17 Sep 2003, 23:08
praet this what u are doing wrong.
7C4 = number of ways of slecting 4 people (includes groups with ben & ann together and also with ann alone & ben alone)
5C2 = groups with ben & ann togther
2*5C3= groups with only ben or ann
Thus groups with neither ann or ben = 7C4 - 5C2 -2*5C3 = 5 (5C4)
hope this helps...
-Vicks
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17 Sep 2003, 23:10
errata: read the last line of solution as:
Thus groups with neither ann nor ben = 7C4 - 5C2 -2*5C3 = 5 (5C4)
-vicks
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17 Sep 2003, 23:55
Vicky wrote:
errata: read the last line of solution as:
Thus groups with neither ann nor ben = 7C4 - 5C2 -2*5C3 = 5 (5C4)
-vicks

i edited this post.

NEITHER .....NOR ...doesnt it mean BOTH

Thats why i did not consider the cases where ben and ann are selected seperately

Praetorian
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18 Sep 2003, 01:46
neither ann nor ben means none of them.
1) What u found out using: 7C4 - 5C2
gives that at most one is present but not both.

2) what 7C4 - 5C2 -2*5C3 gives is the number of selections where none of them neither ann nor ben is present.

3) what 7C4 - 5C4 will give is the number of selections where atleast one of ben or ann is present.

i hope this clarifies. be alert and clear in such questions with what is really being asked...
-Vicks
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18 Sep 2003, 03:22
Vicky wrote:
neither ann nor ben means none of them.
1) What u found out using: 7C4 - 5C2
gives that at most one is present but not both.

2) what 7C4 - 5C2 -2*5C3 gives is the number of selections where none of them neither ann nor ben is present.

3) what 7C4 - 5C4 will give is the number of selections where atleast one of ben or ann is present.

i hope this clarifies. be alert and clear in such questions with what is really being asked...
-Vicks

I totally understand now.....I didnt use the information properly.
in other words... how many combinations if Ben and Ann cannot be selected?

My solution will be valid if the question asks : if Ben and Ann cannot be in the same group,what are the total possible combinations ?

Praetorian
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18 Sep 2003, 15:39
praetorian123 wrote:
Vicky wrote:
neither ann nor ben means none of them.
1) What u found out using: 7C4 - 5C2
gives that at most one is present but not both.

2) what 7C4 - 5C2 -2*5C3 gives is the number of selections where none of them neither ann nor ben is present.

3) what 7C4 - 5C4 will give is the number of selections where atleast one of ben or ann is present.

i hope this clarifies. be alert and clear in such questions with what is really being asked...
-Vicks

I totally understand now.....I didnt use the information properly.
in other words... how many combinations if Ben and Ann cannot be selected?

My solution will be valid if the question asks : if Ben and Ann cannot be in the same group,what are the total possible combinations ?

Praetorian

Sorry guys, I couldn't respond to this sooner. Good discussion, Praet and Vicks! Thanks!
18 Sep 2003, 15:39
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