SOLUTION: how many positive odd numbers can be formed from the digits 2,3,4,5,6 if no digit is to be repeated in a given number 48 120 130 None

I took it that you meant to use ALL 5 digits.  The other tutor took it that
you want 1-digit numbers, 2-digit numbers, etc., all the way to 5-digit
numbers.
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Any odd number ends with an odd digit.

We choose the most restrictive digit first, which is the last digit.  The
last digit can only be one of these two digits 3 and 5. So.

We can choose the last digit any of 2 ways.  That will leave 4 unchosen
digits.

For each of the 2 ways we can then choose the last digit, we can choose the
first digit as any of the 4 remaining unchosen digits.

So we can choose the last and first digits any of 2x4 ways. (8 ways)

That leaves 3 unchosen digits.

For each of the 2X4 or 8 ways we can choose the last and first digits, we
can choose the second digit as any of the 3 remaining unchosen digits.

So we can choose the last, first and second digits as any of 2x4x3 ways.
(24 ways)

That leaves 2 unchosen digits.

For each of the 2X4x3 or 24 ways we can choose the last, first and second 
digits, we can choose the third digit as either of the 2 remaining unchosen
digits.

So we can choose the last, first, second and third digits as any of 2x4x3x2
ways.  (48 ways)

That leaves only 1 unchosen digit. We must choose that final fourth digit as
that 1 remaining unchosen digit.  We can choose it only 1 way. So the final
answer is 2x4x3x2x1 = 48 ways

Edwin