if a women in a restaurant  ,she has five coins (one is (1-cent coin) (one (5-cent coin), (one(10-cent coin),(one(25cent coin), (one(50-cent coin) in how many ways can she leave some (at least one)of her coins  for tip if (a) there are no restriction.

(b) she want to have some chang left.
asked Nov 29, 2016 in Other Math Topics by marya

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There is one set of 5 coins; 5 sets of 4 coins, 10 sets of 3, 10 sets of 2 and 5 sets of one.

1+5+10+10+5=31. This is (a). There is one other set, the null set, which is not used, because she is going to leave at least one coin as a tip. The set consists of 5 coins. The other sets are subsets.

If she wants change left she cannot use the single set of 5, but all the other sets would leave her with at least one coin in change. So (b) is 30. So the sets counted in (b) are a subset of those counted in (a).

answered Nov 30, 2016 by Rod Top Rated User (486,780 points)
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