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Find the number of distinguishable ways the word "STATISTICS" can be arranged if only $1$ T will be alone while the other $2$ T will be together.

How do I solve this? Or does it need complex workings?

I Have done many practices on permutation and Combination. But all I see all the time is questions such as "Number of ways letter T come together" or Number of ways no $2$ Ts are together which also means T cannot be together.

Therefore, the question above is come out by my own self. How do we attempt it?

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We set aside the $T$s for now and arrange the seven letters $S, S, S, I, I, A, C$. We can fill three of the seven positions with an $S$ in $\binom{7}{3}$ ways, fill two of the remaining four positions with an $I$ in $\binom{4}{2}$ ways, and fill the remaining two positions with an $A$ and a $C$ in $2!$ ways. Hence, there are $$\binom{7}{3}\binom{4}{2}2! = \frac{7!}{3!4!} \cdot \frac{4!}{2!2!} = \frac{7!}{3!2!}$$ distinguishable arrangements of the letters $S, S, S, I, I, A, C$.

Arranging these seven letters creates eight spaces in which we can place a TT and a T, six between successive letters, and two at the ends of the row.
$$\square S \square A \square S \square S \square I \square C \square I \square$$ To ensure that exactly two $T$s are together, we must place $TT$ in one of these eight spaces and $T$ in another. Hence, the number of arrangements of the letters of the word $STATISTICS$ in which exactly two $T$s are together is $$8 \cdot 7 \cdot \binom{7}{3}\binom{4}{2}2! = 23520$$

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    $\begingroup$ thanks for your very clear explanation!! i struggle with understanding this topic as textbooks aren't so detailed. So thanks ! $\endgroup$ – mutu mumu Nov 24 '18 at 12:03
  • $\begingroup$ can i check why cannot we use $_8C_2$ but instead must use $8 \cdot 7$ is it because TT and T are "different groups" ? $\endgroup$ – mutu mumu Nov 24 '18 at 12:29
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    $\begingroup$ That is because TT and T are distinguishable, so it matters in which position we place the TT and T. Switching them produces a different arrangement. $\endgroup$ – N. F. Taussig Nov 24 '18 at 12:32
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Here, as I interpret it, $2$ $Ts$ are together, apart from remaining $T$, i.e. all $3$ are not together.

The $10$ letters are:

$$S,S,S,C,A,I,I,T,T,T$$

Put $2$ $T's$ in a box, i.e. put them together, and give it a new name, say t.

So, now there are $9$ letters:

$$S,S,S,A,C,I,I,T,t$$

$S$ and $I$ are repeated thrice and twice resp. Total ways to arrange these $9$ letters is: $$\frac{9!}{3!\cdot2!}$$ Out of this, you will subtract those cases where $t$ and $T$ are together.
So, now assume your letters as:

$$S,S,S,A,C,I,I,Tt$$

Ways to arrange these $8$ letters is: $$\frac{8!\cdot 2!}{3!\cdot 2!}$$

Finally, subtract: $$\frac{9!}{3! \cdot 2!}-\frac{8! \cdot 2!}{3! \cdot 2!}=\frac{7\cdot 8!}{3! \cdot 2!}=23520$$

So, there are $26880$ possible ways it can be arranged, in the given constraints.

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  • $\begingroup$ Your count is wrong. To see why, replace $TT$ by $\tau$, then subtract those cases in which $\tau$ and $T$ are together. $\endgroup$ – N. F. Taussig Nov 24 '18 at 11:52
  • $\begingroup$ @N.F.Taussig isn't it what i did? $\endgroup$ – idea Nov 24 '18 at 11:55
  • $\begingroup$ No. Try the strategy that I suggested, using $\tau$ and $T$, to see why you over counted. $\endgroup$ – N. F. Taussig Nov 24 '18 at 11:56
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    $\begingroup$ @N.F.Taussig Got it! i can arrange t and T in 2 ways... ;-} thanks! $\endgroup$ – idea Nov 24 '18 at 11:58
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    $\begingroup$ Typesetting tip: To create a bold $t$ in math mode, type $\boldsymbol{t}$. $\endgroup$ – N. F. Taussig Nov 24 '18 at 12:07
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Suppose we combine two of the T's into a single character (TT). Then we have 9 characters to arrange, so taking into account the three S's and the two I's, the total number of arrangements is $$N = \frac{9!}{3! 2!}=30240$$ The only problem is that we have included arrangements with three successive T's in this count. In fact we have included each such arrangement twice, once as T(TT) and once as (TT)T. How many such arrangements are there? Consider the three T's to be a single character (TTT). Then the total number of arrangements with TTT together is $$M = \frac{8!}{3! 2!} = 3360$$ So the total number of arrangements in which exactly two T's are together is $$N - 2M = \boxed{23520}$$

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