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# Another simple but tricky one

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Senior Manager
Joined: 06 Apr 2005
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Another simple but tricky one [#permalink]  06 Apr 2005, 17:13
Hi again,

Another question. Bill purchases an item and receives no change. Before the purchase, he had only a five-dollar bill, two ten-dollar bills, and a twenty-dollar bill. How many distinct possibilities are there for the total amount of his purchase.

a) 3
b) 4
c) 6
d) 9
e) 10

Darth
Senior Manager
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Location: USA
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Kudos [?]: 25 [0], given: 1

Ok!! No replies so far. I will wait for a day, before giving the answers
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
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Kudos [?]: 217 [0], given: 0

I moved this topic so that you can get replies
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Best Regards,

Paul

Manager
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Location: NYC
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Re: Another simple but tricky one [#permalink]  07 Apr 2005, 05:19
Hi again,

Another question. Bill purchases an item and receives no change. Before the purchase, he had only a five-dollar bill, two ten-dollar bills, and a twenty-dollar bill. How many distinct possibilities are there for the total amount of his purchase.

a) 3
b) 4
c) 6
d) 9
e) 10

Darth

D
Bills - 5, 10, 10, 20

Possibilities:
5,10,15,20,25,30,35,40,45
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 181 [0], given: 2

9 possibilities

5, 10, 15, 20, 25, 30, 35, 40, 45
Senior Manager
Joined: 19 Feb 2005
Posts: 487
Location: Milan Italy
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yes D for me too
SVP
Joined: 03 Jan 2005
Posts: 2250
Followers: 13

Kudos [?]: 216 [0], given: 0

Re: Another simple but tricky one [#permalink]  07 Apr 2005, 08:49
C(3,1)+C(3,2)+C(3,3)+2
=3+3+1+2=9
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Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Last edited by HongHu on 08 Apr 2005, 07:53, edited 1 time in total.
Senior Manager
Joined: 06 Apr 2005
Posts: 355
Location: USA
Followers: 1

Kudos [?]: 25 [0], given: 1

The prices are mulitples of $5, between$5 and $45. i.e. - 5, 10, 15, 20, 25, 30, 35, 40, 45 Good to see all of you getting the right answers. Darth Senior Manager Joined: 15 Mar 2005 Posts: 421 Location: Phoenix Followers: 2 Kudos [?]: 12 [0], given: 0 [#permalink] 08 Apr 2005, 01:37 Any solution other than a brute force? I tried approaching it with a (Total Bills) ! / (num of$5 bills)! x (num of $10 bills) x (num of$20 bills)

approach, but even that has a problem that two different solutions can yield the same sum (like 2 $5 bills is same as 1$10 bill).

Anyone has a more elegant solution?
_________________

Who says elephants can't dance?

SVP
Joined: 03 Jan 2005
Posts: 2250
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Kudos [?]: 216 [0], given: 0

See my post above.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Senior Manager
Joined: 15 Mar 2005
Posts: 421
Location: Phoenix
Followers: 2

Kudos [?]: 12 [0], given: 0

HongHu wrote:
See my post above.

How about some explanation to go with it Hong?
_________________

Who says elephants can't dance?

SVP
Joined: 03 Jan 2005
Posts: 2250
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Kudos [?]: 216 [0], given: 0

Re: Another simple but tricky one [#permalink]  08 Apr 2005, 11:19
HongHu wrote:
C(3,1)+C(3,2)+C(3,3)+2
=3+3+1+2=9

From the five, the ten and the twenty you could pick one, two and three, each would produce a distinct sum. Then you could also pick both the ten and the twenty, and both the ten, the twenty, and the five.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Manager
Joined: 15 Feb 2005
Posts: 247
Location: Rockville
Followers: 1

Kudos [?]: 6 [0], given: 0

I vote d too
just listed it out from 5 to 45
Manager
Joined: 17 Dec 2004
Posts: 73
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Kudos [?]: 6 [0], given: 0

i got 9 as well, by listing out the combinations of bills.
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