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.

Learn more about quantities in https://brainly.com/question/11993520

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