We would like to test 3 different treatments on a particular type of plant. A worker at the local greenhouse will allow us to use 3 amoung 10 plants.
*Note: Each treatment will be used on one and only one plant.*
a) In how many different ways can we assign the plants to the treatments?
Can i answer: by permutations... 10 by 9 by 8 = 720 ways?
I think I answered this question yesterday, but let's try it again. I may have gotten it wrong since I see each treatment is used only once.
There are 10 choose 3 ways to select 3 plants. Since there are 3 possible treatments that can be used, there are 3 choices for the first plant, 2 for the second and only 1 for the third.
This means the total ways is
3!*(10 choose 3) ways the plants can be chosen and the treatments applied.
We would use permutations for the applications and combinations for the way the plants are chosen.
10 choose 3 is 10!/[3!*7!]
3!*10!/[3!*7!]=10*9*8=720 would be sufficient for you answer.
Yes, you are correct. The number of ways we can assign the plants to the treatments is indeed 720. By using the formula for combinations, which is 10 choose 3 (10C3), we can calculate that there are 120 different ways to choose 3 plants out of 10.
Then, we multiply this by the number of ways we can assign the treatments to the chosen plants. Since each treatment is used only once, we have 3 choices for the first plant, 2 choices for the second plant, and 1 choice for the third plant. Therefore, the total number of ways to assign the plants to the treatments is 3! (3 factorial), which is equal to (3 x 2 x 1) = 6.
Finally, we multiply the number of ways the plants can be chosen (120) by the number of ways the treatments can be assigned (6), giving us a total of 720 different ways to assign the plants to the treatments.
So, the answer to the question is 720, which can be obtained by using permutations and combinations.