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Hello marathonrunner, Thanks for the post! 1. It says Bed A must contain 5 flowers or shrubs, and Bed A should have exactly 1 shrub and exactly 2 annuals, meaning that we have 2 places left. Since we only have 3 choices (annuals, perennials, shrubs), perennials will be our only choice (we cannot add annuals because it says we can only have 2, and we can't add shrubs because it says we can only have 1) 2. For Bed B, we are required to have 2 shrubs and AT LEAST 1 annual, again, 2 places left. But this time is different, because the number of annual can go from 1 up to 3 (we cannot add 4 annuals because the total will be more than 5), and the number of perennials change as we change the number of annuals. We need to discuss all 3 situations. If we add 1 annual (meaning that we add 2 perennials): we choose 1 from 6 types of annual: 6, and we choose 2 from 4 types of perennials: 4*3 / 2=6 (choose 1 from 4, then choose 1 from 3 left, divided by 2 because we cannot use them repeatedly. That's how I calculated, you may have your own way), times them together we get 36. For the remaining 2 situations, we do the same thing. Hope my answer help!
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Hi , what exactly does this mean "No flower or shrub will used more than once in each bed" ? This made me think to use permutation as order matters (no flowers or should not be used more than once). Surely am missing something.
For Bed A, we’re told that the gardener must include exactly 1 type of shrub and exactly 2 types of annual flower. The problem also specifies that bed A must contain a total of 5 different types of shrubs or flowers, so bed A must also contain 2 types of perennial flower.
To choose 1 shrub from 7 possibilities, we calculate 7!/(1!6!) = 7. When choosing only 1, there is always the same number of possibilities as there is of items (7 possibilities, so 7 choices). To choose 2 annual flowers from 6 possibilities, we calculate 6!/(2!4!) = 15. To choose 2 perennial flowers from 4 possibilities, we calculate 4!/(2!2!) = 6. There are 7 possibilities for a shrub, 15 possibilities for the annual flowers, and 6 possibilities for the perennial flowers. In total, there are 7×15×6 = 630 possibilities for bed A.
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