SOLUTION: How many ways are there to create a 6 digit pin number where exactly 4 of digits are the same? the numbers 0-9 can be in any place
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Question 1104768: How many ways are there to create a 6 digit pin number where exactly 4 of digits are the same? the numbers 0-9 can be in any place Found 3 solutions by Alan3354, greenestamps, math_helper:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! How many ways are there to create a 6 digit pin number where exactly 4 of digits are the same? the numbers 0-9 can be in any place
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0 to 9 for the 4 digits are equal --> 1 of 10 choices.
The remaining 2 are 1 of 10 each.
10*10*10= 1000 PINs
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It is 10800.
I was close.
I have a completely different interpretation of the problem, and a completely different answer, than the other tutor....
(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.
The total number of different PINs is 10*36*30 = 10800.
You can put this solution on YOUR website! I think Alan3354's answer does not account for the cases where the identical digits are separated by one or two of the other two digits (i.e. the 4 identical digits do not behave exactly like a single digit).
I had 10800 written in my notebook but I hadn't worked out the problem for the case where you pick the 2 not-necessarily-identical digits first, so seeing 'greenstamps' solution reminded me to try it:
If one picks the 2 not-necessarily-identical digits first: C(10,2) = 45 ways
Then the 4 identical digits can be picked in C(8,1) = 8 ways
And the arrangements are 6!/4! = 30 ways
45*8*30 = 10800 which agrees with the selection done the other way (greenstamps answer)