Answer: 6.
Step-by-step explanation: We can solve this problem by using the combination formula:
C(n, r) = n! / (r! * (n - r)!)
In this case, n is the number of choices (the digits 2, 3, 4, 5, 6, and 9) and r is the number of items in each combination (4 digits).
Plugging in the values, we get:
C(6, 4) = 6! / (4! * 2!) = (6 * 5 * 4 * 3) / (4 * 3 * 2) = 30
Therefore, there are 30 four-digit odd numbers that can be formed using the digits 2, 3, 4, 5, 6, and 9. However, not all of these numbers will be less than 5000, so we need to further filter the list to only include those that meet this requirement.
The four-digit odd numbers that can be formed using these digits are:
2359, 2395, 2539, 2593, 2935, 2953, 3259, 3295, 3529, 3592,
3925, 3952, 5239, 5293, 5329, 5392, 5923, 5932, 9235, 9253,
9352, 9523, 9532
Out of these numbers, only 2359, 2539, 2935, 3925, 5239, and 9523 are less than 5000. Therefore, there are 6 four-digit odd numbers less than 5000 that can be formed using the digits 2, 3, 4, 5, 6, and 9.
Therefore, the final answer is 6.