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Another simple but tricky one

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Another simple but tricky one [#permalink] New post 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

Answer in 8 hours time.

:evil:
Darth
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 [#permalink] New post 07 Apr 2005, 03:36
Ok!! No replies so far. I will wait for a day, before giving the answers
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 [#permalink] New post 07 Apr 2005, 03:50
I moved this topic so that you can get replies
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Re: Another simple but tricky one [#permalink] New post 07 Apr 2005, 05:19
Darth_McDaddy wrote:
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

Answer in 8 hours time.


:evil:
Darth




D
Bills - 5, 10, 10, 20

Possibilities:
5,10,15,20,25,30,35,40,45
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 [#permalink] New post 07 Apr 2005, 06:20
9 possibilities

5, 10, 15, 20, 25, 30, 35, 40, 45
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 [#permalink] New post 07 Apr 2005, 07:15
yes D for me too :twisted:
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Re: Another simple but tricky one [#permalink] New post 07 Apr 2005, 08:49
C(3,1)+C(3,2)+C(3,3)+2
=3+3+1+2=9
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Last edited by HongHu on 08 Apr 2005, 07:53, edited 1 time in total.
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And the answer is... [#permalink] New post 08 Apr 2005, 01:29
Answer is D 9.

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.

:evil:
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 [#permalink] New post 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?
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 [#permalink] New post 08 Apr 2005, 07:54
See my post above. ;)
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 [#permalink] New post 08 Apr 2005, 11:12
HongHu wrote:
See my post above. ;)


How about some explanation to go with it Hong? :)
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Re: Another simple but tricky one [#permalink] New post 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.
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 [#permalink] New post 08 Apr 2005, 15:56
I vote d too
just listed it out from 5 to 45
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 [#permalink] New post 10 Apr 2005, 06:18
i got 9 as well, by listing out the combinations of bills.
  [#permalink] 10 Apr 2005, 06:18
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