binomial theorem: proving equations

Question

Can someone show me how to prove $^{12}C_6$=$(^6C_0)^2+(^6C_1)^2+(^6C_1)^2+(^6C_3)^2+(^6C_4)^2+(^6C_5)^2+(^6C_6)^2$? Thanks so much!

Answer 1

Write it as $$^6C_0\times ^6C_6 + ^6C_1\times ^6C_5 + \ldots + ^6C_6 \times ^6C_0$$ And view this as the number of ways to choose a set of size $6$ from two distinguished subsets of size $6$ (making $12$ altogether).

Answer 2

Hint:

Expand $(1+x)^{12}=(1+x)^6(1+x)^6$ and calculate the coefficient of $x^6$, using that $\;\dbinom nk=\dbinom n{n-k}$.

More generally you can prove that $$\sum_{k=0}^n\binom nk^2=\binom{2n}n.$$

Answer 3

$^nC_r = \frac{n!}{r!(n-r)!}$

Then $LHS = {12\choose 6} = \frac{12!}{6!6!} = \frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7}{6\cdot 5\cdot 4\cdot 3\cdot 2}$ $=\frac {11\cdot 3\cdot 8\cdot 7}{2} = 11\cdot 3\cdot 4\cdot 7=924$

$RHS = {6\choose 0}^2+{6\choose 1}^2+{6\choose 2}^2+{6\choose 3}^2+{6\choose 5}^2+{6\choose 6}^2 = 1+6^2+(\frac{6!}{2\cdot 4!})^2+(\frac{6!}{3!3!})^2+6^2+1$

$=2+72+\frac{6^2\cdot 5^2}{4}+\frac{6^2\cdot 5^2\cdot 4^2}{6^2} = 74+225+400 =924$

$\implies RHS=LHS $