The value of $$\binom{10}{0}^{2}-\binom{10}{1}^{2}+\binom{10}{2}^{2}......-\binom{10}{9}^{2}+\binom{10}{10}^{2}$$ is:
A)$\binom{10}{5}$
A)$\binom{10}{5}^{2}$
c)$-\binom{10}{5}$
d) $10!$
Asked
Active
Viewed 38 times
-1
onelessproblem
- 383
-
It's the coefficient of $x^{10}$ in $\left(\sum_{n=0}^{10}\binom{10}n x^n\right)\left(\sum_{n=0}^{10}(-1)^n\binom{10}n x^n\right)$. – Angina Seng Jun 16 '17 at 06:04
-
If this were of any real concern, it would not be all that difficult to evaluate the sum and compare the result to the multiple choice of answers. When a problem is quoted so precisely without any indication that its meaning has been digested, Readers may ask the OP to add context to the Question. – hardmath Jun 16 '17 at 15:47
1 Answers
2
Actually the it's the coefficient of $x^{10}$ in $$(x+1)^{10}(1-x)^{10}=(1-x^2)^{10}$$ which is $$\binom{10}5(-1)^5$$
lab bhattacharjee
- 274,582