How many derangements of {1, 2, 3, 4, 5, 6} begin with the integers 1, 2, and 3, in some order?

Total derangement = 4

Step-by-step explanation:

To find - How many derangement of {1, 2, 3, 4, 5, 6} begin with the integers 1, 2, and 3, in some order?

Solution -

Given that,

The first three numbers are 1, 2, 3

The only derangement possible are - 231 and 312

Now,

The remaining elements 4, 5, 6 are the 4th, 5th and 6th number in the derangement

Now,

In order for the permutation of the derangement, the only possibility for 123  are 564 and 645

∴ we get

The Total derangement are -

231564

231645

312564

312645

So,

Total derangement = 4

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