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in a window display at a flower shop, there are 3 spots for 1 plant each. to fill these 3 spots, adam has 7 plants to select from, each of a different type. Selecting from the 7 plants, adam can make how many display arrangements with 1 plant in each spot? (Note: the positions of the unselected plants do not matter.)

Respuesta :

JeanaShupp

Answer: 210

Step-by-step explanation:

We know that the number of combinations of n things taken r at a time is given by :-

[tex]C(n,r)=\dfrac{n!}{r!(n-r)!}[/tex]

So, number of ways to select 3 plants out of 7 = [tex]C(7,3)=\dfrac{7!}{3!4!}=\dfrac{7\times6\times5\times4!}{6\times 4!}=7\times5=35[/tex]

Also number of ways to arrange them in 3 positions = 3! = 6

Now , total number of arrangements with 1 plant in each spot = (number of ways to select 3 plants out of 7) x (number of ways to arrange them in 3 positions)

= 35 x 6

=210

Hence, required number of ways = 210

fichoh

Since, the arrangement does not matter, we use the principle of combination, Hence, the number of display arrangement possible is 210

  • Total number of plants = 7
  • Number of spots = 3

This could be defined thus as : (7C3) × 3! ;

3! = number of ways to arrange the 3 plants into the 3 spots

Recall :

  • nCr = [tex] \frac{n!}{(n-r)!r!} [/tex]

7C3 = [tex] \frac{7!}{(7-3)!3!} = \frac{7!}{4!r3!} [/tex]

7C3 = [tex] \frac{7×6×5}{6} = 35 [/tex]

3! = 3 × 2 × 1 = 6

(7C3) × 3! = 35 × 6 = 210 ways

Therefore, there are 210 possible arrangements which can be made.

Learn more : https://brainly.com/question/14767366

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