Ben Brink
Posts:
198
From:
Rosenberg, TX
Registered:
11/11/06


RE: Permutation problems
Posted:
May 22, 2012 1:56 PM



Dr. Scales: This site is serving as a de facto tutoring service for higherlevel students who otherwise would be hardpressed to get any help. Professionals like you make that possible. Thanks. Ben
> Date: Sun, 20 May 2012 15:25:22 0400 > From: discussions@mathforum.org > To: discretemath@mathforum.org > Subject: Re: 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? > > > No strings can contain (fish + bird) or ( bird + rat), so the key is finding how many contain (fish + rat). > > I proceeded as follows: > Start with (fish). The initial letter can be in 23 positions, P1 to P23. > With initial letter in P1, this leaves 22 spaces for (rat), so rat can be in 20 different positions, each leaving 19! permutations of the other letters. > > In P2 we get 19 x 19!, P3 18 x 19!, P4 18 x 19! > P5 to P20 give 16 cases of 18 x 19! > P21, P22 P23 give respectively 18, 19 and 20. > > So the total is 19!.(20+19+18+18+16.18+18+19+20)= 19!.420 > > As a check put (rat) in positions P1 to P24, and (fish) in available positions. > > Now we get 19!.(20+19+18+17+16.17+17+18+19+20)=19!.420 > > So the final answer is 26!23!24!23!420.19! > > > > 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. > > > With 987 in first 3 positions there are 7! permutations, > with 123 in last 3 positions, likewise 7! permutations, > with 45 in 2 fixed positions there are 8! permutations, > so the total is 7!+8!+7!=10.7! > > Regards, Peter Scales.

