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3 years ago

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there are 14 NBA teams who do not make the playoffs of these teams 3 of them will be randomly selected to make the 1st, 2nd, and

3rd pick. how many different ways can the 1st-3rd pick be arranged

2 answers:

mihalych1998 [28]3 years ago

7 0

Answer:

In 2184 different ways can the 1st-3rd pick be arranged.

Step-by-step explanation:

We are given the Total number of NBA teams does not make the play offs = 14

We need to find number of ways in which 3 teams randomly picked.

We use permutation to find the number of ways.

We know that number of ways of selecting r item from n different item is equal to $^nP_r\:\:=\:\:\frac{n!}{(n-r)!}$

Here, r = 3 and n = 14

$\implies^{14}P_3=\frac{14!}{(14-3)!}frac{14!}{11!}=frac{14\times13\times12\times11!}{11!}=14\times13\times12=2184$

Therefore, In 2184 different ways can the 1st-3rd pick be arranged.

monitta3 years ago

6 0

The 1-3 picks could be arranged almost 25 different times

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By comparison

$\mathbf{I_x + 12 = 36.\ I_y + 6 =30}$

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$\mathbf{I = (24,24)}$

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So, we have:

$\mathbf{I = (\frac{E_x + B_x}{2},\frac{E_y + B_y}{2})}$

This gives

$\mathbf{(24,24) = (\frac{18 + B_x}{2},\frac{15 + B_y}{2})}$

Multiply through by 2

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By comparison

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So, we have:

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Hence, the coordinates of B are:

$\mathbf{B = ( 30,33 )}$

Read more about midpoints at:

brainly.com/question/18068617

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