Suppose that a "code" consists of 4 digits, none of which are repeated. (a digit is one of the 10 numbers 0,1,2,3,4,5,6,7,8,9) h ow many codes are possible?
1 answer:
The answer is 5040. This is permutation without the repetition: P = n!/(n - r)! n - the number of choices r - the number of chosen We have: n = 10 r = 4 P = 10!/(10 - 4)! = 10!/6! = (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / (6 * 5 * 4 * 3 * 2 * 1) = 10 * 9 * 8 * 7 = 5040
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