A jar contains 5 quarters, 2 nickels, and 4 pennies. In how many ways
can you arrange all coins in a row so that each arrangement:(1) begins with a quarter?
(2) all quarters together?
(3) two nickels are not next to each other?
(4) no 2 pennies are next to each other?
I believe I solved the first question. It would be 5 * (10!/(4!4!2!)) which equals 3150.
For (2), I got 7!/(4!2!) = 105
For (3) I got the "total – 2 nickels together" approach. So (11!/(5!4!2!)) – (10!/(5!4!)) = 5670.
I'm still very stuck on #4.
It seems to be a MISSISSIPPI problem.
Best Answer
$(4)$ No two $P\; \text{Pennies}$ are not next to each other, If we arrange all $P$ in between $2$ other letters
So Arrangements as $_Q_Q_Q_Q_Q_N_N_$
So total arrangements as $\displaystyle = \underbrace{\frac{7!}{5!\times 2!}}_{\bf{arrange\; all \; Q\; and \; N}}\times \underbrace{\binom{8}{4}}_{\bf{arrange \; 4\;P\; in \; 8\; gap}} = \frac{7\times 6}{2\times 1}\times\frac{8\times 7 \times 6 \times 5}{4\times 3 \times 2 \times 1}$
So we get Answer is $ = 42\times 35 = 1470$