Arrange 'ARRANGED' s.t. A and N aren't next to each other

Question

I was looking at a solution for a question regarding permutations at this thread Arranging letters with two letters not next to each other and I thought of another question.

In how many ways can the word ARRANGED be arranged if A and N aren't next to each other ?

Do I need to take into account that there are 2 As ? Since I would need to find

Number of ways to arrange - Number of ways to arrange such that A and N are next to each other.

This is currently what I have tried:

(1) Number of ways to arrange - $\frac{8!}{2!2!} = 10080 $

(2) Number of ways to arrange st. A and N are next to each other - $ 2*\frac{7!2!}{2!2!} = 5040 $

(3) Therefore, number of ways to arrange the st. A and N aren't next to each other = 5040.

I multiplied (2) by 2 to take into account that either As can be beside N. Would this be correct?

Answer 1

Use inclusion/exclusion principle:

• Include the total number of arrangements: $\frac{8!}{2!2!}=10080$
• Exclude the number of arrangements containing AN: $\frac{7!}{2!}=2520$
• Exclude the number of arrangements containing NA: $\frac{7!}{2!}=2520$
• Include the number of arrangements containing ANA: $\frac{6!}{2!}=360$

Hence the number of arrangements without A and N next to each other is:

$$10080-2520-2520+360=5400$$

Answer 2

How many ways can $\rm RRGED$ be arranged

How many ways can $\Box\Box\Box\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare$ be arranged such that.

• No white square are adjacent.
  • then how many ways can the white square be filled with $\rm AAN$
• Exactly two white square are adjacent
  • then how many ways can these white squares be filled with $\rm AAN$ such that the two adjacent white squares don't contain $\rm AN$
• Don't bother counting ways exactly three white square are adjacent
  • none of these are safe.

Put it all together.

Answer 3

Complications are arising because $N$ can be at the ends or in the middle, and there are $2 A's$ to contend with, but theses complications can be handled rather neatly using stars and bars

$NRRGED$ can be permuted in $6!/2 = 360$ ways, and whether $N$ is at an end or not, there will always be five "boxes" in which the two $A's$ can be placed any which way, as shown below.

$NR\boxed.R\boxed.G\boxed.E\boxed.D\boxed.\;\; or \;\;\boxed.RNR\boxed.G\boxed.E\boxed.D\boxed.\;,$ with the two $A's$ placed in $\binom{2+5-1}{5-1} = 15$ ways

Thus answer $=360\cdot15 = 5400$

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Added another method for those not knowing "stars and bars"

Denoting $N$ by a crossed ball among white ones, an arrangement without the $A's$ would be like

$\huge{\circ\circ\oplus\circ\circ\circ:\;}$ $RRNGED\;6!/2 = 360 \;ways$

Wherever the $N$ might be, the first $A$ would have $5$ places for insertion

$\huge\uparrow\circ\uparrow\circ\oplus\circ\uparrow\circ\uparrow\circ\uparrow:\;$ $360\cdot 5\; ways$

and wherever the first $A$ was inserted (black ball), the second $A$ would have $6$ places for insertion

$\huge\uparrow\circ\uparrow\bullet\uparrow\circ\oplus\circ\uparrow\circ\uparrow\circ\uparrow:\;$ $360\cdot5\cdot6/2\; ways$

Answer 4

We first place the $N$, then the $A$'s, and then the rest.

If $N$ occupies one of the two end cells there are $6$ cells left for the $A$'s, and if $N$ occupies one of the six inner cells there are $5$ cells left for the $A$'s. I all there are $$2{6\choose 2}+6{5\choose2}=90$$ admissible placements for these three letters. For each such placement the remaining $5$ letters $RRGED$ can then be placed arbitrarily in the $5$ remaining cells in ${5!\over2!}=60$ ways.

It follows that there are $90\cdot60=5400$ admissible arrangements in all.