Inclusion/Exclusion Combinatorics

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Homework Statement



Determine the number of permutations of {1,2,3,4,5,6,7} in which exactly four integers are in there natural positions.

The Attempt at a Solution



Would this be solved by using the Inclusion/Exclusion Principle and finding

<br /> \left|S\right| - \sum \left|A_{1}\right| + \sum \left|A_{1} \cap A_{2}\right| -...<br />

where i={1,2,3,4,5,6,7}
 
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Wait the easier way to do this would be to say there are

<br /> \begin{pmatrix}7\\4\end{pmatrix}<br />

while the other 3 integers will form a derangement, so the answer would be

<br /> \begin{pmatrix}7\\4\end{pmatrix} D_{3}<br />

Is that correct?
 
Yes. What is your value for D3?
 
Came out to 2 so my answer was 70
 
It seems good to me.
 
Is a non-recursive formula a formula that only works for a given function and parameters?
 
What do you mean given function and parameters? As far as I know, a non-recursive formula is one not defined in terms of previous terms.
 
Well the question states as follows:

The sequence f_n is defined by f_0=f_1=2 and

f_n = (\frac{f_{n-1}+2f_{n-2}}{6}), when n\geq2.

Find a non-recursive formula for f_n
 
It's asking you to find a formula for fn only in terms of n, not fanything.
 
  • #10
so you need to only find f as a function of n alone
 
  • #11
So its the same as finding a recurrence relation?
 
  • #12
What you gave in the problem is the recurrence relation. You need to find what is called a "closed form." From Wikipedia: Solving a recurrence relation means obtaining a closed-form solution: a non-recursive function of n.
 
  • #13
Shoot, not really sure how to do that...
 
  • #14
would it be the same as this problem:

solve the recurrence relation h_n = 5h_{n-1}+6h_{n-2}

subject to the initial values h_0 = 1 and h_1 = -2
 
  • #15
It is similar enough that the same solution technique would likely work on both.
 
  • #16
I will attempt that and post if I run into trouble. Thanks.
 
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