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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/ 
Associated Topics:
High School Permutations and Combinations

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