Respuesta :
Answer:
The value is [tex]Z = 1120 \ eight -digit \ code [/tex]
Step-by-step explanation:
From the question we are told that
The birth date of Shannon is 12/21/1982
Generally the number of digits present in her birth date is N = 8
Looking at the date we can see that 1 repeated 3 times
2 repeated 3 times
Generally the total number eight -digit code that she could make using her birthday given the total number of digits that make up her birth date is N = 8
is mathematically represented as
K = 8!
but but given that digit 1 repeated itself 3 times , and 2 repeated itself 3 times then the total number of eight -digit code that she could make using her birthday will be mathematically evaluated as
[tex]Z = \frac{K}{3 ! * 3!}[/tex]
=> [tex]Z = \frac{8!}{3 ! * 3!}[/tex]
=> [tex]Z = \frac{8* 7*6*5* 4 * 3!}{3 ! * 3* 2*1 }[/tex]
=> [tex]Z = 1120 \ eight -digit \ code [/tex]
Shannon could make 40,320 eight-digit codes using her digits in her birthday.
Given that Shannon was born on 12/21/1982, to determine how many eight-digit codes could she make using the digits in her birthday, the following calculation must be performed:
- 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = X
- 56 x 30 x 12 x 2 = X
- 1680 x 24 = X
- 40320 = X
Therefore, Shannon could make 40,320 eight-digit codes using her digits in her birthday.
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