Finite help! Combinatorics: I need help with this question! I've tried it multiple times but I don't think I came up with the right answer

Question

Using digits from this set {1,2,3,4,5,6,7}, how many different 3 digit numbers can be formed if digits cannot be repeated and the 3 digit must be odd?

 

I came up with 36 as an answer and I don't think it is the right one

Answer 1

The choices may be made from 1,3,5,7 four odd, 2,4,6 three even

The first two choices contain 0 odd numbers, may be made in 3×2=6 ways, followed by 4 choices of an odd end =24

The first two choices contain 1 odd number, may be made in

4×3=12 ways odd, even or, 3×4=12 ways even, odd total =24 ways , followed by 3 choices of an odd end =72

The first two choices contain 2 odd numbers, may be made in 4×3=12 ways, followed by 2 choices of an odd end =24

Total number of such numbers is 24+72+24=120.  Apologies for the prior error.

 

The total number of 3 digit numbers formed without repetition from 1,2,3,4,5,6,7 is 7×6×5=210

 

 

 

Answer 2

Total 3 digit numbers is 7! / 4! = 210

The number is odd if it ends in one of the odd digits. if the number starts with an odd digit, there are 3 odd numbers for the final digit. If the number starts with an even number, there are 4 numbers for the final digit. So the fraction of 3 digit numbers that begin with odd and end with odd is 210(4/7)(1/2) = 60.  The fraction of 3 digit numbers that begin with even and end with odd is 210(3/7)(2/3) = 60 So 60 + 60 = 120

your answer 36 is the number of 3 digit combinations if the order doesn't matter so perhaps you used the expression 7!/3!4! ?