We have four coins
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Coin 1: $0.10
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Coin 2: $1.00
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Coin 3: $1.00
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Coin 4: $1.00
How many ways can we get $20.00 from these coins?
My attempt:
I started by counting the total number of ways for each coin to reach $20.00
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200 ways for coin 1
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20 ways for coins 2, 3, and 4.
I now have an equation, a + b + c + d + e = 200
We want to get the total number of solutions without any constraints
$\dbinom{200+5-1}{5-1} = \dbinom{204}{4}$
Then I found the number of solutions with the constraint that the coin must be $\leq$ 20. The total number of solutions with coin > 20 can be determined by purchasing 21 coins and leaving 181 at most to buy.
$\dbinom{181+5-1}{5-1} = \dbinom{185}{4}$
The final solution is:
$$\dbinom{204}{4} – 3 \times \dbinom{185}{4} $$
Best Answer
There should be a multiple of 10 number of coins of type $1$. Hence, in effect we have four types of coin of \$ $1$ and desire non-negative solutions to the equation $x_1+x_2+x_3+x_4=20$, which by the stars and bars problem is $\binom{23}{3}$.