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

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

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