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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.

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$(4)$ 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$

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  • $\begingroup$ Juan, thanks for the help. I'm still not entirely clear. The answer should be 1470. $\endgroup$ – borderlineNovice Sep 24 '15 at 3:55
  • $\begingroup$ To borderlineNovice My answer is Right, If you calculate it you will get exact answer. $\endgroup$ – juantheron Sep 24 '15 at 4:01
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$(1)$ Why multiply by 5 ? Quarters are identical.

$(2)$ Ok.

Proceed .....

$(4)$ You can permute the $7$ {quarters $+$ nickels} in $\dfrac{7!}{5!2!}$ ways

They have 8 gaps (including ends) where pennies allowed to be placed, so multiply by $\dbinom{8}{4}$ to get the answer of $1470$

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  • $\begingroup$ Multiply by 5 as there are 5 Qs that can possibly be put there. $\endgroup$ – borderlineNovice Sep 24 '15 at 3:48
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    $\begingroup$ As I said, quarters are identical, so you don't need the multiplier. Imagine there are only 4 quarters + others. Arrange them, and just place a quarter at the beginning. $\endgroup$ – true blue anil Sep 24 '15 at 3:56
  • $\begingroup$ Thanks so much for your help! :) $\endgroup$ – borderlineNovice Sep 24 '15 at 6:18
  • $\begingroup$ You're welcome | $\endgroup$ – true blue anil Sep 24 '15 at 6:55

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