Ben Brink
Posts:
196
From:
Rosenberg, TX
Registered:
11/11/06


RE: Permutation problems
Posted:
May 18, 2012 8:48 AM



Rahesh, Will check on your details later. But assuming you have correct answers to A through C, you may be able to use the pigeonhole principle: n(A u B u C) = n(A) + n(B) + n(C)  n(A n B)  n(A n C)  n(B n C) + n(A n B n C), where "u" denotes "union" and "n" denotes "intersection." But first check over your work on A, B and C. Thanks and will try to look at this later; it's a great problem! Ben
> Date: Fri, 18 May 2012 03:46:00 0400 > From: discussions@mathforum.org > To: discretemath@mathforum.org > Subject: Permutation problems > > Q1. How many permutations are there of the 26 letters of the english alphabet that do not contain any of the strings fish,rat or bird? > > I was able to solve it partially. > A: Permutations that do not contain fish > B: Permutations that do not contain rat > C: Permutations that do not contain bird > > A=26!23! (i found 23! permutations that contain fish) > B=26!24! > C=26!23! > > Stuck at this point. Could not find intersection of A,B,C > > > Q2. How many permutations of 10 digits either begin with the three digits 987,contain the digits 45 in the fifth and sixth positions or end with the three digits 123. > > A: Begin with 3 digits 987, containd 45 in the fifth and sixth positions > B: End with 3 digits 123 > > A=3!*5!*2!=1440 > B=3!*7!=30240 > A intersection B=3!*3!*2!=72 > > A union B=30888. I get this answer but this isn't correct. Please find the error. > > > Thank You.

