lucy was born on 01/01/1966. how many eight - digit codes could she make using the digits in her birthday?
lucy was born on 01/01/1966. how many eight - digit codes could she make using the digits in her birthday?
Answer
Explanation:
Step1: Count the number of digits and repeated digits
The birthday is 01/01/1966. The digits are (0,1,0,1,1,9,6,6). There are (n = 8) digits in total. The digit (0) is repeated (r_1=2) times, the digit (1) is repeated (r_2 = 3) times, and the digit (6) is repeated (r_3=2) times.
Step2: Use the formula for permutations of multi - sets
The formula for the number of permutations of a multi - set with (n) objects, where there are (k) types of objects with (n_1,n_2,\cdots,n_k) repetitions respectively is (\frac{n!}{n_1!n_2!\cdots n_k!}). Here (n = 8), (n_1=2) (for (0)), (n_2 = 3) (for (1)), (n_3=2) (for (6)). [ \begin{align*} \frac{8!}{2!3!2!}&=\frac{8\times7\times6\times5\times4\times3!}{2\times1\times3!\times2\times1}\ &=\frac{8\times7\times6\times5\times4}{2\times2}\ &=\frac{6720}{4}\ & = 1680 \end{align*} ]
Answer:
(1680)