It is a question from permutations and combination chapter and its ans is 48 as given in book! Please help me to do this. I am unable to figure out the solution! Please help!
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To be even it ends with an even digit. Half of the digits are even and it's equally likely to end with one digit as another so 1/2 of them are even and half are odd. So the are 6! Numbers so 6!/2=$123453=360$ are even. Or we could work out the are 3 choices for last digit. 5,4,3,2,1 for the next so $354321$ total. – fleablood Sep 19 '17 at 06:47
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The answer is badly wrong. If you have to use all the digits you are asking for the number of permutations of your digits with the last one being even. You have $3$ choices for the units digit, then $5$ for the tens digit, $4$ for the hundreds and so on. You get $\frac {6!}2=360$ choices. If you don't have to use all the digits you have even more choices.
Ross Millikan
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Hint: you can tell whether or not a number is even by looking solely at its last digit. Try conditioning on the last digit and counting each case individually.
TomGrubb
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