In how many ways can 9 men stand in row if A and B cannot stand side by side, and C must stand at the end?

Along the same lines: How many different permutations can be made of the letters of the word "dependent"?





I'm lost on whether to use Pn,r or Cn,r or a combination of them. Any pointers and explanations would be appreciated.

In how many ways can 9 men stand in row if A and B cannot stand side by side, and C must stand at the end?
Here's my solution





C standing at the end reduces the problem to 8 men





So there are 8! permutations, including A %26amp;B standing together of arranging 8 men.





Consider A and B to effectively be one person





So the number of possibilities with A %26amp; B together is 7!





The number of permutations where A%26amp; B do not stand together = total number – number where they *do* stand together:





So there are 8! – 7! = 8*7! – 7! = (8 – 1)*7! = 7*7! permutations where A %26amp;B do not stand together





Finally, now add C to each permutation: C can stand at either end, so the total number of permutations is





2*7*7! = 70,560








The number of permutations made from the letters of 'dependent' is a different calculation





N = 9!/(3!*2!*2!) = 15,120





The 9! Comes from the total number of letters.





The 3! comes from the number of permutations from three e's since each 'e' is indistinguishable. This reduces the number of permutations by a factor of 3!





Similarly the two d's and the two n's reduce the number of permutations by a factor of 2! twice.





Hope that makes sense.