h3.JPG
For part C) The book has given the answer of:
= 7! x 4! + 8! x 3! - 5! x 4! x 3!
= 345 600
I think this is incorrect, according to my working, the answer is
Working:
Probability of potted roses adjacent
= 7! x 4! (Book gets it right)
+ Probability of azaleas adjacent:
= 8! x 3! (Book gets it right)
- Probability of potted roses and azaleas adjacent:
h5.JPG
= 4! x 4! x 3! x 2
Total = 7! x 4! + 8! x 3! - 4! x 4! x 3! x 2
= 355968
So is the book incorrect?
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It seems like your question comes down to: "what is the number of ways of having the roses adjacent and also the azaleas adjacent?" (since you agree with the book on the other components)
Label the plants as R1, R2, R3, R4, A1, A2, A3, X, Y, Z. (The Rj's stand for roses and Aj's for azaleas.) Then write [R] for a chunk of four adjacent roses and [A] for three adjacent azaleas. In order to have both the roses adjacent and the azaleas adjacent, we need an arrangement like
X, [R], [A], Z, Y.
To make these arrangements, we
• arrange the above (5! ways);
• in each arrangement of the above, arrange the roses in [R] and azaleas in [A] (4!*3! ways).
Hence there are 5!*4!*3! ways to have the roses adjacent and also the azaleas adjacent, which agrees with the book.
Last edited by InteGrand; 12 Feb 2018 at 10:57 PM.
what happens to the probability of both
When you calculated the no. of ways to have roses adjacent and azaleas adjacent, you also made the block of roses adjacent with the block of azaleas. However, this is not necessary in order to have roses adjacent and azaleas adjacent. The meaning of "roses adjacent and azaleas adjacent" in this situation is that there is a block [R] somewhere and also a block [A] somewhere.
Last edited by InteGrand; 12 Feb 2018 at 11:03 PM.
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