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June 2015 p63 q7
2846
Rachel has 3 types of ornament. She has 6 different wooden animals, 4 different sea-shells and 3 different pottery ducks.
She lets her daughter Cherry choose 5 ornaments to play with. Cherry chooses at least 1 of each type of ornament. How many different selections can Cherry make?
Solution
To solve this problem, we need to consider all possible combinations where Cherry chooses at least one of each type of ornament. The types are wooden animals (W), sea-shells (S), and pottery ducks (D).
We can break down the combinations as follows:
1 W, 1 S, 3 D: The number of ways to choose 1 wooden animal is 6, 1 sea-shell is 4, and 3 pottery ducks is \(\binom{3}{3} = 1\). Total = \(6 \times 4 \times 1 = 24\).
1 W, 3 S, 1 D: The number of ways to choose 1 wooden animal is 6, 3 sea-shells is \(\binom{4}{3} = 4\), and 1 pottery duck is 3. Total = \(6 \times 4 \times 3 = 72\).
3 W, 1 S, 1 D: The number of ways to choose 3 wooden animals is \(\binom{6}{3} = 20\), 1 sea-shell is 4, and 1 pottery duck is 3. Total = \(20 \times 4 \times 3 = 240\).
1 W, 2 S, 2 D: The number of ways to choose 1 wooden animal is 6, 2 sea-shells is \(\binom{4}{2} = 6\), and 2 pottery ducks is \(\binom{3}{2} = 3\). Total = \(6 \times 6 \times 3 = 108\).
2 W, 1 S, 2 D: The number of ways to choose 2 wooden animals is \(\binom{6}{2} = 15\), 1 sea-shell is 4, and 2 pottery ducks is \(\binom{3}{2} = 3\). Total = \(15 \times 4 \times 3 = 180\).
2 W, 2 S, 1 D: The number of ways to choose 2 wooden animals is \(\binom{6}{2} = 15\), 2 sea-shells is \(\binom{4}{2} = 6\), and 1 pottery duck is 3. Total = \(15 \times 6 \times 3 = 270\).
Adding all these combinations gives the total number of selections:
