Quantity of an object after proliferation A small creature called "Charza" lives in blocks of an infinite square. An infinite number of them can live in a block. After one hour , one Charza is divided into 4 Charzas and each one moves into one of the adjacent blocks. we have only one Charza at the first . after 6 hours , how many Charzas will be in a block which has only one common vertex with the first block? ( for example , after one hour , there will be one Charza in each adjacent block )

 A: Impose a coordinate system that makes the originally occupied cell $\langle 0,0\rangle$. Let $c(m,n,t)$ be the number of Charzas in cell $\langle m,n\rangle$ at time $t$ hours; clearly $c(0,0,0)=1$, $c(m,n,0)=0$ for $\langle m,n\rangle\ne\langle 0,0\rangle$, and the recurrence is 
$$c(m,n,t+1)=c(m-1,n,t)+c(m+1,n,t)+c(m,n-1,t)+c(m,n+1,t)\;.$$
It’s not hard to see that $c(m,n,t)=c(|m|,|n|,t)$ for all $m,n,t\in\Bbb Z$. 
A little numerical experimentation suggests that if we set
$$d(m,n,t)=\frac{t-(|m|+|n|)}2$$
for $m,n\in\Bbb Z$ and $t\in\Bbb N$, then
$$c(m,n,t)=\begin{cases}
\binom{t}{d(m,n,t)}\binom{t}{m+d(m,n,t)},&\text{if }|m|+|n|\equiv t\pmod 2\\\\
0,&\text{otherwise}\;.\end{cases}\tag{1}$$
Clearly $(1)$ satisfies the initial conditions on $c$, so it only remains to show that it satisfies the recurrence. Suppose that $m,n\ge 0$ and $m+n\equiv t+1\pmod2$. Then
$$\begin{align*}
\binom{t+1}{d(m,n,t+1)}&=\binom{t+1}{\frac12(t+1-m-n)}\\\\
&=\binom{t}{\frac12(t-1-m-n)}+\binom{t}{\frac12(t+1-m-n)}
\end{align*}$$
and
$$\begin{align*}
\binom{t+1}{m+d(m,n,t+1)}&=\binom{t+1}{m+\frac12(t+1-m-n)}\\\\
&=\binom{t}{m+\frac12(t-1-m-n)}+\binom{t}{m+\frac12(t+1-m-n)}\;,
\end{align*}$$
so
$$\begin{align*}
&c(m,n,t+1)=\\\\
&\binom{t}{\frac12(t-1-m-n)}\binom{t}{m+\frac12(t-1-m-n)}+\\\\
&\qquad\binom{t}{\frac12(t-1-m-n)}\binom{t}{m+\frac12(t+1-m-n)}+\\\\
&\qquad\binom{t}{\frac12(t+1-m-n)}\binom{t}{m+\frac12(t-1-m-n)}+\\\\
&\qquad\binom{t}{\frac12(t+1-m-n)}\binom{t}{m+\frac12(t+1-m-n)}=\\\\
&\\\\
&\binom{t}{d(m,n+1,t)}\binom{t}{m+d(m,n+1,t)}+\\\\
&\qquad\binom{t}{d(m+1,n,t}\binom{t}{m+1+d(m+1,n,t)}+\\\\
&\qquad\binom{t}{d(m-1,n,t)}\binom{t}{m-1+d(m-1,n,t)}+\\\\
&\qquad\binom{t}{d(m,n-1,t)}\binom{t}{m+d(m,n-1,t)}=\\\\
&\\\\
&c(m,n+1,t)+c(m+1,n,t)+c(m-1,n,t)+c(m,n-1,t)\;,
\end{align*}$$
as desired.
Added: The specific number desired is
$$c(1,1,6)=\binom62\binom63=15\cdot20=300\;.$$
A: sorry , I was very busy . 
My solution : after n hours ,we write the number of walks from center to a block . so we should count the number of walks from $(0.0)$ to $(1,1)$ and this is equivalent to counting the number of sequences of length 6 , consisted of elements called U , R , D , L _( R = L + 1 ) , ( U = D + 1 ) and the sum of these elements should be 6 . so we have 3 situations:
$$ U = 3 , D = 2 , R = 1 , L = 0 $$
$$ U = 2 , D = 1 , R = 2 , L = 1 $$
$$ U = 0 , D = 1 , R = 3 , L = 2 $$
so $\frac{6!}{0!1!2!3!} + \frac{6!}{1!2!1!2!} + \frac{6!}{2!3!1!0!} = 60 + 180 + 60 = 300$
