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

 Messages: [ Previous | Next ]
 Peter Scales Posts: 192 From: Australia Registered: 4/3/05
Re: Permutation problems
Posted: May 20, 2012 3:25 PM

> 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.

Date Subject Author
5/18/12 Mahesh
5/18/12 Ben Brink
5/19/12 Mahesh
5/19/12 Ben Brink
5/19/12 Angela Richardson
5/19/12 Ben Brink
5/20/12 Peter Scales
5/22/12 Ben Brink
8/11/13 Mark Rickert
8/11/13 Ben Brink
8/12/13 Mark Rickert
8/12/13 Ben Brink
8/12/13 Peter Scales
8/12/13 Ben Brink