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In how many ways can the letters of the english alphabet be arranged so that there are seven letter between the letters A and B, and no letter is repeated?

I have searched this question and have seen many interpretations.

Answer-1 (source: answers.yahoo.com)

Ans: 24 x 23 x 22 x21 x20 x19 x 18

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Answer-2: (source: m4maths.com)

Ans 24P7 * 2 * 18

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Answer-3: (source: careerbless) Ans: 24!*36

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My solution

since there are 36 positions for A and B (1-9, 9-1, 2-10,10-2, ...18-26,26-18) and remaining 24 letters can be positioned in 24! ways, to me, the third answer 24!*36 appears right. Please tell if this is correct.

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  • $\begingroup$ In your solution you meant 36 positions for A and B. $\endgroup$ – user21820 Mar 2 '16 at 9:54
  • $\begingroup$ @user21820 thx, it was a typo, corrected $\endgroup$ – Kiran Mar 2 '16 at 9:55
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Correct. 18 positions for the pair (A,B), and 2! ways to arrange them in those positions, and 24! ways to put the remaining 24.

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A slightly different approach, just to confirm your answer:

  • Choose a place for A and put B appropriately ahead: $26-(7+1)$ options
  • Reorder A and B in every possible way: $2!$ options
  • Choose $7$ letters out of the letters between C and Z: $\binom{26-2}{7}$
  • Reorder those $7$ letters in every possible way: $7!$ options
  • Reorder the remaining letters in every possible way: $(26-2-7)!$ options

The answer is therefore: $(26-(7+1))\cdot2!\cdot\binom{26-2}{7}\cdot7!\cdot(26-2-7)!=24!\cdot36$

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