SOLUTION: How may different 5-digit numbers that are odd and less than 50,000 can be formed using the digits 2,3,4,5,6,7 and 8 without repetition?

Question 87547: How may different 5-digit numbers that are odd and less than 50,000 can be formed using the digits 2,3,4,5,6,7 and 8 without repetition?
Found 3 solutions by checkley75, kapheine, sudhanshu_kmr:
Answer by checkley75(3666)   (Show Source):
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THE UNITS DIGIT MUST BE ODD & YOU HAVE 3 TO CHOSE FROM.
THE 10'S DIGIT CAN BE ANY ONE OF THE REMAINING 6 DIGITS.
THE 100'S DIGIT CAN BE ANY ONE OF THE REMAINING 5 DIGITS.
THE 1,000'S DIGIT CAN BE ANY ONE OF THE RAMAINING 4 DIGITS.
THE 10,000'S DIGIT CAN BE ANY OF THE DIGITS 2,3 OR 4 = 3 DIGITS.
THUS WE HAVE 3*6*5*4*3=1080 NUMBERS.

Answer by kapheine(1)   (Show Source):
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I disagree with the previous presented solution, because the "3" cannot be used both in the first and the last digit, the answer 1080 is too high..
This is what I have:
(for each first digit being) -- (next 3 digits combinations) * (last digit)
2 -- 6C3 * 3C1
3 -- (special case) 6C3 * 2C1
4 -- 6C3 * 3C1
add these 3 categories together and we get a total of 960!

Answer by sudhanshu_kmr(1130)   (Show Source):
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At unit place only one of these are possible 3,5,or 7.

Case A (when 3 at unit place i.e fifth position) :
for first position i.e _xxxx only 2 or 4 is possible otherwise it will greater
than 50,000

so, no. of ways to fill first position = 2 (two digits have used )

now, no. of ways to fill second position = 5 ( only 5 digits are remaining)

no. of ways to fill third position i.e xx_xx = 4 (only 4 digits are remaining)

no. of ways to fill fourth position = 3 (only 3 digits are remaining)
fifth position is already occupied by 3.
so, total no. of different numbers that can be formed = 2 * 5 * 4 * 3 =120
..........................................................................

Case B (when 5 at unit place i.e fifth position) :
for first position i.e _xxxx only 2, 3 or 4 is possible otherwise it will
greater than 50,000
so, no. of ways to fill first position = 3 (two digits have used )
now, no. of ways to fill second position = 5 ( only 5 digits are remaining)
no. of ways to fill third position i.e xx_xx = 4 (only 4 digits are remaining)
no. of ways to fill fourth position = 3 (only 3 digits are remaining)
fifth position is already occupied by 5.
so, total no. of different numbers that can be formed = 3 * 5 * 4 * 3 =180

...............................................................................
Case C (when 7 at unit place i.e fifth position) :
similarly,
so, no. of ways to fill first position = 3 (two digits have used )
now, no. of ways to fill second position = 5 ( only 5 digits are remaining)
no. of ways to fill third position i.e xx_xx = 4 (only 4 digits are remaining)
no. of ways to fill fourth position = 3 (only 3 digits are remaining)
fifth position is already occupied by 7.

so, total no. of different numbers that can be formed = 3 * 5 * 4 * 3 =180

now we add the value of these three cases.
no. of different 5-digit numbers that are odd and less than 50,000 can be
formed = 120 + 180 + 180 = 480

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