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.)

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

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

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

Recall :

7C3 =

7C3 =

3! = 3 × 2 × 1 = 6

(7C3) × 3! = 35 × 6 = 210 ways

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

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210

Step-by-step explanation:

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

So, number of ways to select 3 plants out of 7 =

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

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.