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In a lottery daily game, a player picks four numbers from 0 to 9 (without repetition). How many different choices does the player have

a) If order matters?


b) If order does not matter?

Respuesta :

SaniShahbaz

Answer:

a) If order matters, choices the player have = 5040

b) If order does not matter, choices the player have = 210

Step-by-step explanation:

n = 10, r = 4

When the order matters, its permutation.

permutation without repetition = P(n, r) =  [tex]\frac{n!}{(n - r)!}[/tex]

= [tex]\frac{10!}{(10 - 4)!}[/tex]

= [tex]\frac{10!}{6!}[/tex]

= 5040

When the order doesn't matter, its combination.

combination without repetition= C(n, r) = [tex]\frac{n!}{r!(n - r)!}[/tex]

= [tex]\frac{10!}{4!(10 - 4)!}[/tex]

= [tex]\frac{10!}{4! × 6!}[/tex]

= 210

If order matters choices the player has is [tex]5040[/tex].

A)Apply permutation without repetition when order matters

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

where [tex]n=10(0,1,2,3,4,5,6,7,8,9)[/tex]

[tex]r=4[/tex]

[tex]P(10,4)=\dfrac{10!}{(10-4)!}[/tex]

[tex]P(10,4)=\dfrac{10!}{6!}[/tex]

[tex]P(10,4)=\dfrac{10\times9\times8\times7\times6!}{6!}[/tex]

[tex]P(10,4)=10\times9\times8\times7[/tex]

[tex]P(10,4)=5040[/tex]

B) Apply combination without repetition when order doesn't matters

[tex]C(n,r)=\dfrac{n!}{r!(n-r)!}\\C(n,r)=\dfrac{10!}{4!(10-4)!}[/tex]

[tex]C(n,r)=\dfrac{10!}{4!(6)!}[/tex]

[tex]C(n,r)=\dfrac{10\times9\times8\times7\times6!}{4!(6)!}[/tex]

[tex]C(10,4)=210[/tex].

Choice does the player has if order does not matter is [tex]210[/tex].

Learn more about permutation and combination here ,

https://brainly.com/question/4546043?referrer=searchResults