Asked by Lindz
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.
*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.
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