Calculating number of times exactly 2 pairs of consecutive identical letters.

Question

I am troubled by this problem, my intuition fails literally and I can't seem to understand why.

In how many ways can the letters in ARRANGEMENT be arranged so that there are exactly two pairs of consecutive identical letters?

My approach:

Number of ways at least two such pairs can occur is $4\choose2$*$\frac {9!}{2!*2!}$

Number of ways at least three such pairs can occur is $4\choose3$*$\frac {8!}{2!}$

Number of ways at least four such pairs can occur is $4\choose4$*$7!$

So total number of ways in which exactly 2 pairs occurs should be $4\choose2$$\frac{9!}{2!*2!}$ - $4\choose3$$\frac{8!}{2!}$ + $4\choose4$*$7!$

Answer 1

Your count of the number of ways you can have at least two pairs seems wrong. Perhaps if you explain your calculation, I can help you pinpoint what's wrong.

Let's count, for instance, the number $K$ of permutations in which $AA$ and $RR$ appear, but not $NN$ and not $EE$. (As there are six pairs among A, R, N, E, your final answer will be $6K$.) We have

$$K = (\text{number of perms with AA, RR}) - (\text{number of perms with AA, RR, NN}) - (\text{number of perms with AA, RR, EE}) + (\text{number of perms with AA, RR, NN, EE}).$$

So $K = 9!/(2!)^2 - 2 \cdot 8!/2! + 7! = 11 \cdot 7!.$

Therefore the answer is $6K = 66 \cdot 7! = 332640.$

Answer 2

According to the hypothesis for each $1 \leq m \leq t$ ,the number of elements in $S$ that satisfy exactly $m$ of the conditions $c_1 , c_2 , .., c_t $ given by $$E_m= S_m - C(m+1,1)S_{m+1} + C(m+2,2)S_{m+2} - ... + (-1)^{t-m}C(t , t-m) S_{t} $$

So , by using the hypothesis $$E_2= S_2 - C(3,1)S_{3} + C(4,2)S_{4} $$ where $ S_2= C(4,2) \times 90720 $ , $ S_3= C(4,3) \times 20160 $ , $ S_4= C(4,4) \times 5040$

$S_2$ means the number of all arrangements where $2$ pairs are adjacent among $AA,EE,NN,RR$.

$S_3$ means the number of all arrangements where $3$ pairs are adjacent among $AA,EE,NN,RR$.

$S_4$ means the number of all arrangements where $4$ pairs are adjacent among $AA,EE,NN,RR$.

By calculation answer is $544320 - (3 \times 80640) + (6 \times 5040 ) = 332640$

BE CAREFUL !! This question is different from ordinary principle of inclusion -exclusion , because it is asked for $\color{red}{exactly}$ $2$ pairs . Hence , I used the generalized version of principle of inclusion -exclusion . When you use principle of inclusion -exclusion directly , you make overcounting . To prevent overcounting , we used the given formula. You can draw a venn diagram to see how the formula works.

Answer 3

We shall compute by breaking it into two parts

Glue the two $A's$ and two $E's$ together as $\Bbb A, \Bbb E$ respectively, so we have

$\Bbb A \Bbb E GMTRRNN$ with $\frac{9!}{2!2!} = 90720$ permutations

In part $2$, subtract permutations where neither $R's\; nor\; N's$ are together

Arrangements with $N's$ together = $\frac{8!}{2!}=20160$

Ditto for $R's$ together $= 20160$

Both $N's\;and\; R's$ together $= 7!=5040$

By inclusion- exclusion, neither $R's\;nor N's$ together = $90720- 2*20160+5040=55440$

Thus exactly $A's\; and\; E's$ together $= 55440 $

Finally, there can be $\binom42$ double pairs, giving the answer as $6*55440= \boxed{332640}$