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Topic: Permutation problems
Replies: 13   Last Post: Aug 12, 2013 2:23 PM

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Ben Brink

Posts: 198
From: Rosenberg, TX
Registered: 11/11/06
RE: Permutation problems
Posted: May 22, 2012 1:56 PM
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Dr. Scales:
This site is serving as a de facto tutoring service for higher-level students who otherwise would be hard-pressed 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.





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