Permutations & Combinations
Permutation and Combination
star_batch_jee_advanced_2025
Grade 11
Question:
The number of ways in which 10 students can be divided into three teams, one containing 4 and others 3 each, is:
\frac{10!}{4!3!3!}
2100
^{10}C_4 \times ^5C_2
\frac{10!}{6!3!3!} \cdot \frac{1}{2}
Step-by-Step Solution
Key Concept: The two identical teams of 3 must be accounted for by dividing by $2!$ to prevent counting the same partition twice.
We need to divide 10 students into three teams: one with 4 students and two with 3 students each. First, choose 4 students for the first team: $^{10}C_4$ ways. Then, from the remaining 6 students, choose 3 for the second team: $^6C_3$ ways. The last 3 automatically form the third team. However, since the two teams of 3 are indistinguishable (both have equal size), we divide by $2!$ to avoid overcounting: $ rac{^{10}C_4 imes ^6C_3}{2!} = rac{10!}{4!3!3! \cdot 2}$. This equals $^{10}C_4 imes ^5C_2$ because $^5C_2 = rac{^6C_3}{2!}$. Calculating: $210 imes 10 = 2100$.
Correct Answer: 2,3