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Manager
Joined: 29 May 2008
Posts: 110

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15 Jun 2009, 06:34
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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
Manager
Joined: 08 Feb 2009
Posts: 141
Schools: Anderson

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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.
Manager
Joined: 29 May 2008
Posts: 110

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

How did you get the answer

Manager
Joined: 08 Feb 2009
Posts: 141
Schools: Anderson

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

How did you get the answer

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.
Manager
Joined: 29 May 2008
Posts: 110

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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!!
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