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Here is the question:

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I JUST WANT TO KNOW THE MISTAKE IN MY METHOD:

Here is how I solved but my answer is incorrect: the correct answer is (7!/2!) 8C3 enter image description here Thankyou

Yb609
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    Total number of ways - number of ways with two T's together (combine two T's and treat it as one T). – Yunus Syed Mar 28 '17 at 06:08
  • @YunusSyed ya here I did the same. But unable to reach the answer. Please can you help me out? – Yb609 Mar 28 '17 at 06:10
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    @Yb609, Posting images of your hand written notes is not a decent way of asking questions. You should type whatever image you have uploaded. – Parish Mar 28 '17 at 06:18
  • Not only are images unsearchable and greatly decrease the usefulness of your question to future readers, it also makes your post inaccessible to people with certain neurotypes or disabilities. Your handwriting - like virtually all cursive - is close to illegible to me. – Stella Biderman Mar 28 '17 at 16:44
  • @YunusSyed: This is in principle a good insight, but it should be considered as bringing out a subtlety in the meaning of "at random" concerning how we count the outcomes. Are the two $T$'s (actually three of them) distinguishable (in counting), or not? This ties in with how the event space for "at random" is chosen/defined. – hardmath Mar 28 '17 at 16:57

3 Answers3

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First, replace every letter that's not a $T$ with a $0.$ The number of ways you can arrange the symbols in $0TT000T000$ with no successive $T$s is the same as the number of solutions of the equation $x_1+(1+x_2)+(1+x_3)+x_4=7$ in nonnegative integers $x_1,x_2,x_3,x_4;$ by the so-called "stars and bars" formula that is $\binom83.$

Next we have to replace the seven zeros $0000000$ with letters $AACINOR$ in random order; the number of ways to do that is $\binom7{2,1,1,1,1,1}=\frac{7!}{2!}.$

So the final answer is $$\binom83\cdot\frac{7!}{2!}=\boxed{141120}.$$

Alternatively:

The total number of words we can make with the ten letters $AACINORTTT$ is $$\frac{10!}{2!3!}.$$ From this we have to subtract the number of words in which two (or more) $T$s occur consecutively. Let's glue two of the $T$s together and call the resulting symbol $T^2.$ The number of words we can make with the nine letters $AACINORTT^2$ is $$\frac{9!}{2!}.$$ This is not exactly the number we want to subtract, because words with three consecutive $T$s are counted twice, i.e., $TTT$ is counted both as $TT^2$ and as $T^2T.$ The number of words containing three consecutive $T$s is the number of words we can make with the eight letters $AAT^3CINOR,$ that is, $$\frac{8!}{2!}.$$ Therefore, the number of words containing two or more consecutive $T$s is $$\frac{9!}{2!}-\frac{8!}{2!}$$ and the number of words with no consecutive $T$s is $$\frac{10!}{2!3!}-\frac{9!}{2!}+\frac{8!}{2!}=\boxed{141120}.$$

bof
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  • Ya I understand your method but can you please say what is incorrect with my method? – Yb609 Mar 28 '17 at 06:56
  • Not if I have to read it from an image of your handwritten paper, because you couldn't be bothered to type it up neatly like I did mine. Too much trouble. – bof Mar 28 '17 at 07:33
  • Is your solution the same as Yunus Syed's answer? The mistake in his answer is that the middle term $$-\frac{9!}{2!2!}$$ should be $$-\frac{9!}{2!}.$$ – bof Mar 28 '17 at 07:50
  • Thanks for your support. I just wanted to know why is 8!/2! added in yunus syed's answer? – Yb609 Mar 28 '17 at 08:00
  • Let me write $T^2$ to stand for two $T$s stuck together. In Syed's *corrected* answer (see my comment to Syed's answer), the term $-\frac{9!}{2!}$ does some *double counting*. For instance the arrangement $ATTTRACION$ is counted twice, once as $AT^2TRACION$ and once as $ATT^2RACION.$ The arrangements which have been counted twice are precisely the ones containing three Ts in a row, and there are $8!/2!$ of those. – bof Mar 28 '17 at 08:09
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We want to compute: (Total Number of Ways) - (Number of ways with two T's together) + (Number of ways three T's) because of overcounting. However, the number of ways two T's are together is the same as combining two T's together as one T and then counting the total number of arrangements.

This is equal to $\frac{10!}{2!3!} - \frac{9!}{2!2!} + \frac{8!}{2!}$

Edit: Correct answer is $\frac{10!}{2!3!} - \frac{9!}{2!} + \frac{8!}{2!}$ with reasoning in comments.

Yunus Syed
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  • I too got the same answer but the correct answer is (7!/2!) 8C3 – Yb609 Mar 28 '17 at 06:15
  • Why did you add 8!/2! ? – Yb609 Mar 28 '17 at 07:01
  • The middle term in your answer is wrong. It should be $$\frac{10!}{2!3!}-\frac{9!}{2!}+\frac{8!}{2!}.$$ – bof Mar 28 '17 at 07:53
  • After you combine two T's you don't have $\text{AACINORTT},$ you have $\text{AACINORT}\mathbb T;$ nine letters but only two of them are identical. – bof Mar 28 '17 at 07:57
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A more graphic way of reaching the same correct solution:

Take the three $T$'s and stick a blank to the right of each $T$. (Think Scrabble tiles.)

Take an additional $5$ blank tiles. Lay these $11$ tiles down in a row, left to right. You now have a row of $11$ tiles; there is at least one blank between any two $T$s, and the $11$th tile is blank and can be discarded.

The number of such distinct rows of letters is the number of ways to arrange $3$ identical things (the double tiles) and $5$ other identical things (the single blanks). This can be done in $N_1$ ways:$$N_1 =\frac{8!}{3! \times 5!} $$The remaining $7$ letters (a pair of $A$s and $R, C, I, O, \text{and }N$) can be arranged in $N_2$ ways; $$N_2=\frac{7!}{2!}$$Now, just drop these $7$ remaining tiles into the blanks in the row of $T$s and blanks.

The total number of arrangements, $N$, then is given by:$$N=N_1 \times N_2= 141,120$$

DJohnM
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