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Step 1: Understand the Problem
We have 10 students: S1, S2, …, S10, and we need to divide them into three groups A, B, and C. The conditions are:
Each group (A, B, and C) must have at least one student.
Group C can have at most 3 students.
We want to count the total number of ways to form such groups.
Step 2: Break Down by Number of Students in Group C
Since group C can have at most 3 students, we consider the following cases separately:
C has 1 student
C has 2 students
C has 3 students
Step 3: Case 1 – Group C has 1 student
Select 1 student out of 10 to place in group C. The number of ways to do this is \binom{10}{1} .
That leaves 9 students to be distributed between groups A and B. Each of these 9 students can go to either A or B, giving 2^9 total distributions. However, we must exclude the cases where all 9 go to A or all 9 go to B. That means we subtract 2 from 2^9 . So, the number of ways to form groups A and B is 2^9 - 2 .
Hence, the total number of ways for this case is:
\[
\binom{10}{1} \times (2^9 - 2).
\]
Numerically, \binom{10}{1} = 10 , and 2^9 - 2 = 512 - 2 = 510 , so the product is 10 \times 510 = 5100 .
Step 4: Case 2 – Group C has 2 students
Select 2 students out of 10 to place in group C. The number of ways to do this is \binom{10}{2} .
That leaves 8 students for groups A and B. Similar to the previous case, the number of ways to split these 8 students into A and B (with neither A nor B empty) is 2^8 - 2 .
Hence, the total number of ways for this case is:
\[
\binom{10}{2} \times (2^8 - 2).
\]
Numerically, \binom{10}{2} = 45 , and 2^8 - 2 = 256 - 2 = 254 , so the product is 45 \times 254 = 11430 .
Step 5: Case 3 – Group C has 3 students
Select 3 students out of 10 for group C. The number of ways to do this is \binom{10}{3} .
The remaining 7 students must be split into groups A and B, neither being empty, giving 2^7 - 2 ways.
Thus, the total number of ways for this case is:
\[
\binom{10}{3} \times (2^7 - 2).
\]
Numerically, \binom{10}{3} = 120 , and 2^7 - 2 = 128 - 2 = 126 , so the product is 120 \times 126 = 15120 .
Step 6: Compute the Total
Adding the results from all three cases gives the final total number of ways to form such groups:
\[
5100 + 11430 + 15120 = 31650.
\]
Final Answer
The total number of ways to form the groups under the given conditions is 31650.
