Combinations of Four-Digit PINs with Exactly Three Digits Different
Date: 08/06/2012 at 04:59:56 From: Raffaele Subject: Trying to find the number PIN codes A Personal Identification Number (PIN) consists of four digits in order, each of which may be any one of 0, 1, 2, ..., 9. Find the number of PINs satisfying each of the following requirements: (a) All four digits are different. (b) There are exactly three different digits. Part a) seems obvious: 10_P_4 = 10*9*8*7 = 5040 permutations. Here is the answer for Part b): We can choose the face value of the pair in 10 ways. We can then choose two other different digits in 9_C_2 = 36 ways. The number of distinct ways to arrange two like and two unlike objects is 4!/(2!*1!*1!) = 12, so the total number of 4-digit PINs with exactly three different digits is 12*10*36 = 4320. I am happy with the answer, but why is 10*9_C_2 used in the first instance rather than 10_C_3? I know that numerically they are different but can't justify the use of one over the other. Having three distinct digits implies that the "fourth" digit is the same as one of the other three. Thanks for your help.
Date: 08/06/2012 at 13:13:00 From: Doctor Douglas Subject: Re: Trying to find the number PIN codes Hi Raffaele, Thanks for submitting your question to Ask Dr. Math. The expression 10*9_C_2 works because it describes how to select three objects such that one of the three is special, and the other two are equivalent. In the original problem, the fact that one of the numbers pair up makes that number distinct from the other two digits eventually chosen. You would use 10_C_3 if you were selecting three *equivalent* objects from a larger group of ten, e.g., a six-digit PIN consisting of three pairs of digits. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/
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