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Permutations Puzzle
By Abiskek Tuli on Saturday, March 25, 2000 - 07:35 pm:

To whom it may concern:
Hello, my name is Abishek Tuli and I am a student at Erindale Secondary School in Mississauga, Ontario in Canada. I am writing to ask for help on a Finite math question in the area of permutations.

Question: Find the number of even 5-digit numbers, using digits 0 to 9 with no repetitions, so that 4 must be in the number. If 2 is in the number it must be the last digit.

I would really appreciate a solution to this problem which I am having difficulty with. I hope that you can help. Thank you very much for taking the time to help me.

By John Grindall (Jhg25) on Saturday, March 25, 2000 - 08:44 pm:


First of all I just want to check I've got the question.
45022 isn't allowed (repetition)
56924 isn't allowed (2 is in the wrong place)
65792 isn't allowed (no four)

OK.

It's probably useful to look at different cases in turn-
a) there is a two in the number
b) there isn't a two in the number

In case a) we know that the two must be in the last decimal place and then the number is automatically even, and there are no more twos.
This leaves four places left to be filled with any different numbers out of (0,1,3,4,5,6,7,8,9); exactly one of them being a four.
Can you now solve case a) ??

In case b) it's helpful to consider two subcases-
i) the four is in the last place
ii) the four isn't in the last place
( and we know that there is exactly one four)


Can you now solve cases bi) and bii) ??

NB in case i) we have that the number is even automatically, but in case ii) you need to look only at inserting digits which make the 5-digit number even.

Now, for any number which solves the problem, one and only one of the cases a, bi), bii) will hold so just add them up.


Be sure to reply if there's something you're still not sure about.

John.