Coins question

J

Jaspworld

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Apr 17, 2007
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Gr 12 question. Should be easy.

Suppose 10 coins are to be chosen from an unlimited supply of pennies, nickels, dimes, and quarters. In how many ways can this be done? Assume that two coins of the same kind are indistinguishable.
:D
 
The answer is the number of solutions using non-negative integers to the equation;
\(\displaystyle p + n + d + q = 10\).

The number of ways of putting N identical objects into K different cells is

\(\displaystyle C(N + K - 1,K) = \left( {\begin{array}{c}
{N + K - 1} \\
N \\
\end{array}} \right) = \frac{{\left( {N + K - 1} \right)!}}{{N!\left( {K - 1} \right)!}}\).

Here you have ten identical ones and four different variables.
 
Hello, Jaspworld!

I think I have an approach to this problem . . .

Gr 12 question. . Should be easy. .Really?

Suppose 10 coins are to be chosen from an unlimited supply
. . of pennies, nickels, dimes, and quarters.
In how many ways can this be done?
Assume that two coins of the same kind are indistinguishable.
Click to expand...

Consider the ten coins placed in a row: \(\displaystyle \:O\:O\:O\:O\:O\:O\:O\:O\;O\:O\)


We will insert three "walls" in the row: \(\displaystyle \,O\:O\:O\,|\,O\:O\,|\,O\:O\:O\:O\,|\,O\)
. . This will denote: 3 pennies, 2 nickels, 4 dimes, 1 quarter.


Each of the three walls can be placed in any of 11 positions.

For example: \(\displaystyle \,O\:O\,|\,|\,O\:O\:O\:O\:O\,|\,O\:O\:O\,\) means: 2p, 0n, 5d, 3q

. . . . . . . . . . \(\displaystyle \,|\,|\,O\:O\:O\:O\:O\:O\:O\,|\,O\:O\;O\,\) means: 0p, 0n, 7d, 3q

. . . . . . . . . . \(\displaystyle O\:O\:O\:O\,|\,|\,|\,O\:O\:O\:O\:O\;O\,\) means: 4p, 0n, 0d, 6q

. . . . . . . . . . \(\displaystyle |\,\:O\:O\:O\:O\:O\:O\:O\:O\:O\:O\,|\,|\,\) means: 0p, 10n, 0d, 0q


Since each of the three walls has eleven choices of position,

. . there are: \(\displaystyle \,11^3 \:=\:\L1331\) possible partitions.

 
huh.

pka is right. Since the book uses the method that soroban used, I'd be more interested in learning that method. I'll check to see where Soroban went wrong.

Thanks a lot guys.
 
Re: huh.

Jaspworld said:
check to see where Soroban went wrong
Click to expand...
soroban said:
Consider the ten coins placed in a row: \(\displaystyle \:O\:O\:O\:O\:O\:O\:O\:O\;O\:O\)
We will insert three "walls" in the row: \(\displaystyle \,O\:O\:O\,|\,O\:O\,|\,O\:O\:O\:O\,|\,O\)
This will denote: 3 pennies, 2 nickels, 4 dimes, 1 quarter.
Click to expand...
That represents the only way to select that particular collection of ten coins.
We do not count the number of ways the collection can be arranged.

Arrange three |'s and ten o's: \(\displaystyle \L \frac{13!} {\(10!\)\(3!\)}\)
 
Code: 
10,0,0,0

9,0,0,1
9,0,1,0
9,1,0,0

8,0,0,2
8,0,2,0
8,2,0,0
8,0,1,1
8,1,0,1
8,1,1,0

7,0,0,3
7,0,3,0
7,3,0,0
7,0,2,1
7,2,0,1
7,2,1,0
7,0,1,2
7,1,0,2
7,1,2,0
7,1,1,1

6,0,0,4
6,0,4,0
6,4,0,0
6,0,1,3
6,1,0,3
6,1,3,0
6,0,3,1
6,3,0,1
6,3,1,0
6,0,2,2
6,2,0,2
6,2,2,0
6,2,1,1
6,1,2,1
6,1,1,2

The pattern seems to be \(\displaystyle \L a_n = \sum_{k=1}^n k = \frac{n(n+1)}{2}\)

So if I'm right (Its late, and I'm tired), your answer could also be given by: \(\displaystyle \L 4 \sum_{n=1}^{10} a_n = 2 \sum_{n=1}^{10} n(n+1)\).
 
daon said:
So if I'm right (Its late, and I'm tired), your answer could also be given by: \(\displaystyle \L 4 \sum_{n=1}^{10} a_n = 2 \sum_{n=1}^{10} n(n+1) = 880\).
Click to expand...
No the correct answer is still:
\(\displaystyle \L \frac{{13!}}{{10!\left( {3!} \right)}} = 286.\)
 
You're right, I goofed. It shouldn't have been multiplied by 4, and should have been summed from 1 to 11, since the "patten" above has 11 steps (0 to 10 inclusive for values in the first entry).

Should have been \(\displaystyle \L \sum_{n=1}^{11} a_n = \frac{1}{2}\sum_{n=1}^{11} n(n+1)=286\)

Sorry for the confusion. I also did not realize how old this thread was.
 
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