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A piggy bank contains 50 pennies, 40 nickels, 30 dimes, and 20 quarters.

(1) How many ways are there to choose 10 coins with at least one of each type?

(2) How many ways are there to choose 10 coins with at least 3 nickels but no more than 2 quar- ters?

For (1), I figured that the problem entails A) choosing 4 coins, one from each type and B) then choosing the rest of the 6 coins. So this would give 1 * C(4+6-1, 6) = C(9,6) = 84.

On (2), I'm quite stuck. I don't know how to approach this problem.

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Your solution for $(1)$ is correct, and you can apply the same ideas to $(2)$.

First choose $3$ nickels, then choose the remaining $7$ coins, in $\binom{4+7-1}7=\binom{10}7=120$ ways. But there should be no more than $2$ quarters, so we need to subtract the choices with more than $2$ quarters. First choose $3$ nickels and $3$ quarters, then choose the remaining $4$ coins, in $\binom{4+4-1}4=\binom74=35$ ways. So the desired number is $120-35=85$.

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    $\begingroup$ Completely clear explanation. Thank you. I see how the same logic applies now. :) $\endgroup$ – borderlineNovice Sep 22 '15 at 20:35

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