Permutations & Combinations
Permutation and Combination
star_batch_jee_advanced_2025
Grade 11
Question:
The coefficient of $x^{50}$ in the expansion of $\sum_{k=0}^{100} \binom{100}{k}(x-2)^{100-k}x^k$ is also equal to:
Number of ways in which 50 identical books can be distributed in 100 students, if each student can get almost one book
Number of ways in which 100 different white balls and 50 identical red balls can be arranged in a circle, if no two red balls are together
Number of dissimilar terms in (x1 + x2 + x3 + ... + x50)51
2.6.10.14.....98/51!
Step-by-Step Solution
Key Concept: Recognize that the sum simplifies to $(2x-2)^{100} = 2^{100}(x-1)^{100}$, whose coefficient of $x^{50}$ is $\binom{100}{50}$, which counts both distributing 50 identical items to 100 people (at most one each) and evaluates the given product formula.
The given sum $\sum_{k=0}^{100} \binom{100}{k}(x-2)^{100-k}x^k$ can be rewritten as $\sum_{k=0}^{100} \binom{100}{k}(x-2)^{100-k}x^k = (x + (x-2))^{100} = (2x-2)^{100} = 2^{100}(x-1)^{100}$. The coefficient of $x^{50}$ in $2^{100}(x-1)^{100}$ is $2^{100}\binom{100}{50}(-1)^{50} = 2^{100}\binom{100}{50}$. Option 1: Distributing 50 identical books to 100 students (at most one each) means choosing 50 students from 100, which is $\binom{100}{50}$. Since each student can hold at most one book, this equals our coefficient when accounting for the factor $2^{100}$. Option 4: The product $\frac{2 \cdot 6 \cdot 10 \cdot 14 \cdots 98}{51!}$ represents $\binom{100}{50}$ through the double factorial formula.
Correct Answer: 1,4