how many ways are there to arrange $8$ pennies and $5$ nickels in a How can you arrange a line of $8$ pennies and $5$ nickels, so that no $2$ nickels are next to each other (pennies are indistinguishable and nickels are too)?
Answer: I set nickels and pennies next to each other so I get ${9 \choose 5}*8!*5!=609638400$. Is this correct?

Imagine a "base" Line of ()N(p)N(p)N(p)N(p)N() and you have have to put the remaining 4 pennies into the six () pots.  This is then what others irritatingly call "stars and bars".  
There are four pots in which to put the leftmost of the four pennies.  Call it n where 6 is the left most pot and 1 is the right most pot.  There are n pots to put the second most left penny  and so on.
SO there are $\sum_{n=1}^6\sum_{m=1}^n\sum_{k=1}^m m$ ways to place the pennies.
$\sum_{n=1}^6\sum_{m=1}^n\sum_{k=1}^m k=$
$\sum_{n=1}^6\sum_{m=1}^n \frac{m(m+1)}2=$
$\frac 12 [\sum_{n=1}^6(\sum_{m=1}^nm^2 + \sum_{m=1}^n m)]=$
$\frac 12 [\sum_{n=1}^6(\frac{n(n+1)(2n+1)}{6}+ \frac{n(n+1)}2   ]=$
$\frac 14 [\sum_{n=1}^6(\frac{2n^3 + 3n^2 + n}{3}+ n^2 + n   ]=$
$\frac 14 [\sum_{n=1}^6(\frac{2n^3  }{3}+ 2n^2 + \frac 43 n   ]=$
$\frac 14 [\frac 23\frac{6^2(6+1)^2}4+ 2\frac{6(6+1)(2*6+1)}6 + \frac 43 \frac {6(6+1)}2   ]=$
$\frac 14 [2* 3*7^2+ 2(7)(13) +  4*7   ]=$
$\frac 74 [2* 3*7+ 2*(13) +  4   ]=$
$\frac 74 [72   ]= 7*18  = 126$
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Which as others point out I could have simply solved by ${9 \choose 5}$... oh, well.
