Question 1104768
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I have a completely different interpretation of the problem, and a completely different answer, than the other tutor....<br>
(1) You can choose any 1 of 10 digits to be the one that occurs 4 times: C(10,1) = 10.
(2) You need to choose 2 of the other 9 digits for the other 2 digits of the PIN: C(9,2) = 36.
(3) The number of different ways you can arrange the 4 identical and 2 other digits is (6!)/(4!) = 720/24 = 30.<br>
The total number of different PINs is 10*36*30 = 10800.