Final answer:
There are 240 different 4-digit odd numbers that can be formed using the digits 1, 2, 3, 4, 5, 6, 7, with each digit appearing only once, by ensuring the last digit is odd and using permutations for the remaining places.
Step-by-step explanation:
The question asks about the number of different odd numbers of 4 digits that can be formed with the digits 1, 2, 3, 4, 5, 6, 7, with each digit being distinct. To determine this, we first need to ensure the last digit is odd, to make the whole number odd. There are 4 odd digits (1, 3, 5, 7) that can be used for the last position. Once we've fixed the last digit as an odd one, we have 6 remaining digits to fill the other three places.
For the first digit (thousands place), we cannot have 0 and cannot reuse the last digit, and thus have 5 options left. Once we place a digit in the first position, we have 5 digits left, from which we choose one for the hundreds place (now having 4 options left). For the tens place, we will be left with 4 digits, and we choose one, leaving 3 options. The mathematical calculation is therefore 5 (options for first place) * 4 (options for second place) * 3 (options for third place) * 4 (options for fourth place). Multiplying these together, we get 5 x 4 x 3 x 4 = 240.
Thus, there are 240 different 4-digit odd numbers that can be formed using the digits 1, 2, 3, 4, 5, 6, 7 without repetition.