Combinatorial proof of derangement identity

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July 28th 2010, 07:51 PM
oldguynewstudent

Combinatorial proof of derangement identity

Could someone check out this combinatorial proof? Thanks.

Let $D_{n}$ be the number of derangements of an n-set. Define $D_{0}=1$ and note $D_{1}=0.$

a) Give a combinatorial proof: $D_{n}=(n-1)(D_{n-1}+D_{n-2})$ for $n\geq2$ .

How many derangements of an n-set are there?

One way to count would be $D_{n}=n!\sum_{j=0}^{n}\frac{(-1)^{j}}{j!}$ .

Another way is to consider the n-set by separating out the n-1 numbers greater than 1, one at a time.

Start with 2. There are $D_{n-1}$ derangements of the n-1 numbers in the n-set excluding the number 2. If we pre-pend the number 2 to the beginning of these derangements, we will still get $D_{n-1}$ derangements because the 2 in the first position will not cause any of the numbers to be “fixed”. Now consider the next number in the n-set, 3. We can proceed the same as we did with the number 2 and there will again be $D_{n-1}$ derangements with the number 3 pre-pended to the derangement. In fact, there will be n-1 such derangements. But notice that none of these derangements will have 1 in the second position of the sequence after the numbers have been pre-pended. That is because after we remove the number from the n-set and take the derangement of the n-1 numbers left, 1 can never be in the first position of that derangement. So, if we now go through the same steps again by removing 1 along with n-1 numbers, we can then take the derangements of the n-2 numbers left. Start with 21 and pre-pend 21 to the derangements of the n-2 numbers left just as we did with the single numbers. Each of these derangements will have $D_{n-2}$ derangements and there will be n-1 of these derangements. When we add the second group of derangements to the first group of derangements it is clear that we now have all the derangements of [n]. Therefore we have proved that $D_{n}=(n-1)(D_{n-1}+D_{n-2})$ for $n\geq2.$
July 28th 2010, 11:01 PM
aman_cc

Another way to approach could be

Ignore the last (nth) element.
The remaining (n-1) elements can be de-arranged in D(n-1) ways, once you do that the nth element can interchange place with any of the (n-1) element to give you a de-arrangement of n elements.
Also, out of the (n-1) element we can have an element which holds its poisition, say element 'i'. Other element (remaining (n-2) elements) can be de-arranged in D(n-2) ways. Once this is done to get de-arrangement of 'n' elements you just interchange positions of ith and nth element. Now, i can take values from 1 to (n-1).

You can see immdiately that no more than 1 of the (n-1) elementss can hold its position. So above two exhaust all the cases for D(n). Hence

D(n) = (n-1).D(n-1) + (n-1)D(n-2)
July 29th 2010, 05:13 AM
oldguynewstudent

Thank you so much for your advice. I really just learned combinatorial proofs in the last month, so I am very encouraged that my proof is not wrong. I have a long way to go to make them concise and totally solid, but this is very encouraging.
July 29th 2010, 05:51 AM
aman_cc

I actually haven't checked your proof please. Actually couldn't really follow you line of argument but it seems wrong to me.
July 29th 2010, 02:40 PM
chiph588@

Quote:

Originally Posted by oldguynewstudent

Could someone check out this combinatorial proof? Thanks.

Let $D_{n}$ be the number of derangements of an n-set. Define $D_{0}=1$ and note $D_{1}=0.$

a) Give a combinatorial proof: $D_{n}=(n-1)(D_{n-1}+D_{n-2})$ for $n\geq2$ .

How many derangements of an n-set are there?

One way to count would be $D_{n}=n!\sum_{j=0}^{n}\frac{(-1)^{j}}{j!}$ .

Another way is to consider the n-set by separating out the n-1 numbers greater than 1, one at a time.

Start with 2. There are $D_{n-1}$ derangements of the n-1 numbers in the n-set excluding the number 2. If we pre-pend the number 2 to the beginning of these derangements, we will still get $D_{n-1}$ derangements because the 2 in the first position will not cause any of the numbers to be “fixed”. Now consider the next number in the n-set, 3. We can proceed the same as we did with the number 2 and there will again be $D_{n-1}$ derangements with the number 3 pre-pended to the derangement. In fact, there will be n-1 such derangements. But notice that none of these derangements will have 1 in the second position of the sequence after the numbers have been pre-pended. That is because after we remove the number from the n-set and take the derangement of the n-1 numbers left, 1 can never be in the first position of that derangement. So, if we now go through the same steps again by removing 1 along with n-1 numbers, we can then take the derangements of the n-2 numbers left. Start with 21 and pre-pend 21 to the derangements of the n-2 numbers left just as we did with the single numbers. Each of these derangements will have $D_{n-2}$ derangements and there will be n-1 of these derangements. When we add the second group of derangements to the first group of derangements it is clear that we now have all the derangements of [n]. Therefore we have proved that $D_{n}=(n-1)(D_{n-1}+D_{n-2})$ for $n\geq2.$

Looks fine to me!