Binomial Theorem
Alternating binomial coefficient sum — incorrect statements
MJAT_TS5_P1
Grade 12
Question:
If $S=\binom{10}{0}\binom{20}{10}-\binom{10}{1}\binom{18}{10}+\binom{10}{2}\binom{16}{10}-\binom{10}{3}\binom{14}{10}+\binom{10}{4}\binom{12}{10}-\binom{10}{5}\binom{10}{10}$, then find the INCORRECT statement(s):
A) Number of divisors of $S$ is $10$
B) Sum of digits of $S$ is $6$
C) $S$ is a four-digit number
D) Number of divisors of $S$ is $9$
Step-by-Step Solution
Key Concept: From solution: $S$ = coefficient of $x^{10}$ in $[(1+x)^2-1]^{10}=x^{10}(1+x)^{10}\cdot\frac{(...)}{...}$... Actually $S$ = coefficient of $x^{10}$ in $[(1+x)^2-1]^{10}=(x^2+2x)^{10}=x^{10}(x+2)^{10}$. So $S=2^{10}=1024$.
$S=1024$. A ✗ (11 divisors), B ✗ (digit sum=7), D ✗ (11 divisors). INCORRECT: A, B, D.
Correct Answer: ABD