answer:The study aimed to determine the hospital-based incidence and causes of traumatic spinal cord injury (SCI) in Tehran, Iran. A total of 138 cases of traumatic SCI were identified, with the majority of patients being male and the mean age being 33.2 years. The annual incidence of hospitalized traumatic SCI was found to be 10.5/1,000,000/year. Falls were found to be the leading cause of injury, followed by road traffic crashes. The duration of hospital stay for tetraplegia was longer than for paraplegia. Early surgery was performed for 19% of patients, and the median day of hospitalization for early surgery was less than for late surgery. Prevention strategies should focus on males aged 21-30, falls, and road traffic crashes. More studies are needed to evaluate the incidence of non-hospitalized traumatic SCI patients.

answer:Since there must be at least one member from each faculty, we know that 6 members will come from each of the 6 faculties. This accounts for 6 * 1 = 6 members in the committee. We are left with 4 more members to select for the committee. Let's use the stars and bars method to solve this problem. We have 4 more members to distribute among the 6 faculties. Think of this as having 4 stars that represent the members and 5 bars placed between them to divide these stars into 6 groups. For example: **|*|**||* This would represent two members from the Arts faculty, one from Economics, none from Education, two from Engineering, and one from Science. Notice that each group will at least have one initial member we already counted earlier. Now we just need to find the number of ways to arrange the stars and bars. This can be represented as choosing 4 places (for the stars) among a total of 9 places (4 stars + 5 bars). This can be represented as the number of combinations: C(9,4) = 9! / (4! * (9-4)!) C(9,4) = 9! / (4! * 5!) C(9,4) = (9*8*7*6) / (4*3*2) C(9,4) = 126 So there are 126 possible ways in which the faculties representation can be distributed on the committee with at least one member from each faculty. Note: we assume that there are more than 10 members in each faculty, as stated in the problem.

answer:My apologies, I misunderstood the problem. Let's try to solve it correctly this time. Since we need at least one member from each faculty, let's start by selecting one member from each of the 6 faculties. We are left with 10 - 6 = 4 more members to be selected from any of the faculties. Now, we have 4 members and 6 faculties. We can imagine this as 4 identical balls and 5 dividers to distribute these 4 balls into 6 categories (faculties). We will use the stars and bars method for this distribution. So, the problem is equivalent to arranging the 4 identical balls and 5 dividers in a row. In total, we have 4 + 5 = 9 items to arrange. Now, we need to choose 4 of these 9 positions to place the balls in. The remaining 5 positions will be where the dividers are placed. So, we can represent the number of arrangements as: C(9, 4) = 9! / (4! * (9 - 4)!) C(9, 4) = 9! / (4! * 5!) C(9, 4) = (9*8*7*6) / (4*3*2*1) C(9, 4) = 126 However, this initial calculation doesn't consider that each faculty has initially one representative. To account for the initial members, we should add one member to each faculty after the distribution. In other words, we are overcounting each partition by a factor of 2^6, because in each of the divisions, we've counted every representative as 2. The correct answer should be: 126 * 2^6 = 126 * 64 = 8,064 However, there is one more thing to consider. We have also overcounted the case where all 10 members are from different faculties, which should only be counted once. So, we need to subtract 63 of those overcounted instances (since the total overcounting factor was 64): 8,064 - 63 = 8,001 Therefore, the correct solution would be 8,001 possible ways to distribute the faculties representation with at least one member from each faculty. Apologies for the confusion in my previous answer.

answer:Apologies for the confusion in my previous responses. Let's try to solve the problem correctly this time. We have 6 faculties and need to form a committee of 10 members with at least one member from each faculty. We can use the principle of inclusion-exclusion to solve this problem. First, let's find the total number of ways to form a committee without any restrictions. Since there are 6 faculties and we need to choose 10 members, this can be represented as: C(6+10-1, 10) = C(15, 10) = 15! / (10! * 5!) = 3,003 Now, we need to subtract the cases where at least one faculty is not represented. There are 6 faculties, so we have 6 cases where one faculty is not represented. For each of these cases, we need to choose 10 members from the remaining 5 faculties: C(5+10-1, 10) = C(14, 10) = 14! / (10! * 4!) = 1,001 Since there are 6 cases, we need to multiply this by 6: 6 * 1,001 = 6,006 Now, we need to add back the cases where at least two faculties are not represented. There are C(6, 2) = 15 ways to choose 2 faculties to be excluded. For each of these cases, we need to choose 10 members from the remaining 4 faculties: C(4+10-1, 10) = C(13, 10) = 13! / (10! * 3!) = 286 Since there are 15 cases, we need to multiply this by 15: 15 * 286 = 4,290 Now, we need to subtract the cases where at least three faculties are not represented. There are C(6, 3) = 20 ways to choose 3 faculties to be excluded. For each of these cases, we need to choose 10 members from the remaining 3 faculties: C(3+10-1, 10) = C(12, 10) = 12! / (10! * 2!) = 66 Since there are 20 cases, we need to multiply this by 20: 20 * 66 = 1,320 Now, we need to add back the cases where at least four faculties are not represented. There are C(6, 4) = 15 ways to choose 4 faculties to be excluded. For each of these cases, we need to choose 10 members from the remaining 2 faculties: C(2+10-1, 10) = C(11, 10) = 11! / (10! * 1!) = 11 Since there are 15 cases, we need to multiply this by 15: 15 * 11 = 165 Finally, we need to subtract the cases where at least five faculties are not represented. There are C(6, 5) = 6 ways to choose 5 faculties to be excluded. However, it is not possible to choose 10 members from only one faculty, so there are 0 cases for this. Now, we can apply the principle of inclusion-exclusion to find the total number of ways to form the committee with at least one member from each faculty: 3,003 - 6,006 + 4,290 - 1,320 + 165 = 3,003 - 1,871 = 1,132 So, there are 1,132 possible ways to distribute the faculties representation with at least one member from each faculty. Apologies for the confusion in my previous answers.