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Coins please Help

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Manager
Manager
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Joined: 29 May 2008
Posts: 111

Kudos [?]: 122 [0], given: 0

Coins please Help [#permalink]

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New post 15 Jun 2009, 06:34
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A
B
C
D
E

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I don-t remember how to solve this one

There is a formula that can be applied to this problem but I don-t remmeber how

(2^n) - 1

I have a penny, a nickel,a dime a quarter and a fifty-cent piece. In how many ways may I draw a different sum of money with these coins?

A) 25
B) 31
C) 7
D) 8
E) 64

Kudos [?]: 122 [0], given: 0

Manager
Manager
User avatar
Joined: 08 Feb 2009
Posts: 145

Kudos [?]: 60 [0], given: 3

Schools: Anderson
Re: Coins please Help [#permalink]

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New post 15 Jun 2009, 11:20
TheRob wrote:
I don-t remember how to solve this one

There is a formula that can be applied to this problem but I don-t remmeber how

(2^n) - 1

I have a penny, a nickel,a dime a quarter and a fifty-cent piece. In how many ways may I draw a different sum of money with these coins?

A) 25
B) 31
C) 7
D) 8
E) 64



I think the answer is B:
\(^5C_1\hspace{5} +\hspace{5} ^5C_2\hspace{5} +\hspace{5} ^5C_3\hspace{5} +\hspace{5} ^5C_4\hspace{5} +\hspace{5} ^5C_5\) = 31.

Kudos [?]: 60 [0], given: 3

Manager
Manager
User avatar
Joined: 29 May 2008
Posts: 111

Kudos [?]: 122 [0], given: 0

Re: Coins please Help [#permalink]

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New post 15 Jun 2009, 11:47
It is true , it is B

How did you get the answer

Would you please help me with an explanation?

Kudos [?]: 122 [0], given: 0

Manager
Manager
User avatar
Joined: 08 Feb 2009
Posts: 145

Kudos [?]: 60 [0], given: 3

Schools: Anderson
Re: Coins please Help [#permalink]

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New post 15 Jun 2009, 12:08
TheRob wrote:
It is true , it is B

How did you get the answer

Would you please help me with an explanation?


It is a Combinations problem.

There are \(\hspace{20} ^5C_1 = \frac{5!}{(1!)(5-1)!} \hspace{20}\) ways of selecting One coin.



There are \(\hspace{20} ^5C_2 = \frac{5!}{(2!)(5-2)!} \hspace{20}\) ways of selecting Two coins.



There are \(\hspace{20} ^5C_3 = \frac{5!}{(3!)(5-3)!} \hspace{20}\) ways of selecting Three coins.



There are \(\hspace{20} ^5C_4 = \frac{5!}{(4!)(5-4)!} \hspace{20}\) ways of selecting Four coins.



There are \(\hspace{20} ^5C_5 = \frac{5!}{(5!)(5-5)!} \hspace{20}\) ways of selecting Five coins.



Sum it all up to get 31.

Kudos [?]: 60 [0], given: 3

Manager
Manager
User avatar
Joined: 29 May 2008
Posts: 111

Kudos [?]: 122 [0], given: 0

Re: Coins please Help [#permalink]

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New post 15 Jun 2009, 12:48
I see and now I know how to apply the formula

2^n - 1

2^5 - 1

32 - 1

Thanks!!

Kudos [?]: 122 [0], given: 0

Re: Coins please Help   [#permalink] 15 Jun 2009, 12:48
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