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Combinations

by CaptainOats » Wed Oct 27, 2010 8:51 am
Hej guys,


14. A famer has 10 sheep 4 black and 6 white. He would like to take 3 of the sheep to get shaved. How many different groups of 3 sheep can he select that would have at least one black one?

15. There are 10 books on a book shelf, 5 Math, 3 History and 2 Art. If 2 books are selected at random how many different pairs of 2 different kinds of books can be selected?

Can you provide an answer to these questions which can help to asnwer Combinations-Q. in general?

Cheers
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by sixpointer » Wed Oct 27, 2010 10:59 am
CaptainOats wrote:Hej guys,


14. A famer has 10 sheep 4 black and 6 white. He would like to take 3 of the sheep to get shaved. How many different groups of 3 sheep can he select that would have at least one black one?

15. There are 10 books on a book shelf, 5 Math, 3 History and 2 Art. If 2 books are selected at random how many different pairs of 2 different kinds of books can be selected?

Can you provide an answer to these questions which can help to asnwer Combinations-Q. in general?

Cheers
Possible cases 1B 2W => 4C1*6C2=60
2B 1 W=> 4C2*6C1=48
3B 0 W =4

60+48+4=112 Is it correct Bro?
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by rros0770 » Wed Oct 27, 2010 11:48 am
Setup looks right Six, just looks like that 2nd calc might be a typo.
(4C2) * (6C1) = 36

I'm coming up with the answer as 100 for the sheep question, I solved it with a different method though. This is an at least Combo question, so:

(Total # Possible Combinations with no restrictions) - (Combinations composed of all White sheep)
(10C3) - (6C3)=
(120)- (20) = 100 combos with at least 1 black sheep



15.
Not sure if I interpretted the wording of this question correctly:

1 Math & 1 History book= 5 * 3 = 15
1 Math & 1 Art book = 5* 2 = 10
1 History & 1 Art book = 3 * 2 = 6

Total= 15 + 10 + 6 = 31


You have the OA to that 2nd question?

[/u]
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by limestone » Wed Oct 27, 2010 9:11 pm
Hi,

Question 1

Number of groups of 3 sheep that would have at least one black sheep = total number of groups - Number of groups of 3 sheep that dont have any black sheep.

Number of groups of 3 sheep that dont have any black sheep = Number of groups of 3 sheep that would have all white sheep = 6C3 = 20

Total number of groups : 10C3 = 120

The answer is: 120 - 20 = 100
sixpointer wrote

Possible cases 1B 2W => 4C1*6C2=60
2B 1 W=> 4C2*6C1=48
3B 0 W =4

60+48+4=112 Is it correct Bro?
There's a miscalculation here, bro. 4C2*6C1 = 6 * 6 = 36
The answer must be: 60+36+4 = 100

Question 2


With 2 books selected at random, we can only form : 2 books of different kinds or 2 books of the same kind
rros0770 wrote
Not sure if I interpretted the wording of this question correctly:

1 Math & 1 History book= 5 * 3 = 15
1 Math & 1 Art book = 5* 2 = 10
1 History & 1 Art book = 3 * 2 = 6

Total= 15 + 10 + 6 = 31
I think that your result is correct. And I'll try the indirect method to confirm it.

Total ways = 10C2 = 45

2 maths = 5C2 = 10

2 History = 3C2 = 3

2 Art = 2C2 = 1

The answer = 45 - 10 - 3 - 1 = 31. => Confirmed.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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by fskilnik@GMATH » Thu Oct 28, 2010 12:22 pm
rros0770 wrote: 1 Math & 1 History book= 5 * 3 = 15
1 Math & 1 Art book = 5* 2 = 10
1 History & 1 Art book = 3 * 2 = 6

Total= 15 + 10 + 6 = 31


You have the OA to that 2nd question?

[/u]
Very good solutions, guys!

Let me make a digression for you all... hope you like it.

You don´t need to be insecure in the total calculated if you are sure there were:

(a) No "double-counting´s" (I mean *at least* double-countings...) ... technically speaking, you must assure the events (you counted) are mutually exclusive, that is, the occurence of each one of them implies the non-occurence of the others.

(b) the events considered are Exhaustive, meaning that their reunion covers (contains) the desired event (in this case, this desired event is "choosing 2 books of different kinds")

Having that in mind, "defining" (at your head, not necessarily explicitly, for sure)

Event A = "Choosing 2 books, 1 Math & 1 History"
Event B = "Choosing 2 books, 1 Math & 1 Art book"
Event C = ...

If you are SURE (a) and (b) are respected (and we are sure), you ARE calculating the proper total/answer!!!

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by diebeatsthegmat » Fri Oct 29, 2010 6:03 pm
CaptainOats wrote:Hej guys,


14. A famer has 10 sheep 4 black and 6 white. He would like to take 3 of the sheep to get shaved. How many different groups of 3 sheep can he select that would have at least one black one?

15. There are 10 books on a book shelf, 5 Math, 3 History and 2 Art. If 2 books are selected at random how many different pairs of 2 different kinds of books can be selected?

Can you provide an answer to these questions which can help to asnwer Combinations-Q. in general?

Cheers
14. select 2 from 10 = 10!/2!*8!=120
+ select 1 B and 2W=4*6!/2!*4!=60
+select 2B and 1W=4!/2!*2!*6=36
+select 3B=4
total=100
so the final result=100/12=5/6
15.
select 2 from 10=10!/2!*8!=45
+ select 2 maths= 5!/3!2!=10
or select 2 arts=1
select 2Histories=3
total=14
the final result=45-14/45=31/45

you can do it with another way which is the same as the number 14's solution

select 2 from 10=45 combinations
+ select 1 h and 1 math =15 combinations
+select 1H and 1 art= 3*2=6
+select 1 M and 1 A=5*2=10
so total combinations= 31
the final posibility=31/45
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