Proof of Number of: *permutations of ‘n’ things, taken ‘r’ at a time, when ‘m’ specified things always come together* I read below at many sources

Number of permutations of ‘n’ things, taken ‘r’ at a time, when ‘m’ specified things always come together =$ m!  * (n-m+1) !$

However no one gave the proof.
I reached till this:


*

*First we have choose r out of n: $^nC_r=\frac{n!}{(n-r)!r!}$

*Then choose m out of r: $^rC_m=\frac{r!}{(r-m)!m!}$

*Next arrange m elements : $m! (r-m+1)!$

*Next I have to multiply all above and do cancelation. So I reached to: $\frac{n!(r-m+1)!}{(n-r)!(r-m)!}$


But I don't to get how to proceed to get the given $ m!  * (n-m+1) !$
 A: Order the things that come together: $m!$
Now treat all the block of $m$ things as a single element, you have $n-m+1$ elements to order.
Order the new group of elements: $(n-m+1)!$
So for each order of the $m$ elements you'll have $(n-m+1)!$ orderings, so...
$m!(n-m+1)!$
Ask for further clarification if needed (saying what's not understood)
A: I think the formula you have been given is incorrect - it does not take account or the condition of "arranging r objects". Your approach does take account of this (so it won't match that formula), but I think it doesn't quite match the language, which would usually refer to m specified objects to be kept together.(e.g. "arrange 12 letters chosen from the Roman alphabet, with P,Q,R,S to appear together")
This would work:
*

* Select the m objects: - assuming they are specified, there is only one way to do this.

* Select the rest of the r objects to be arranged: (n-m)!/((r-m)!(n-m+r)!) ways to do this

* arrange the m specified objects: m! ways - this forms a single large object
 
*  arrange the r-m+1 objects consisting of the large object from step 3 and the other (r-m) objects selected: (r-m+1)! ways to arrange these.

Putting it all together: m!(n-m)!(r-m+1)! / ((n-r+m)!(r-m)!) ways to arrange r objects taken from n objects if m specified objects must be (selected and) kept together. 
A: This is permutation question, so objects can come in different order. 


*

*Treat M objects as single entity/object. Remember, inside this entity, m objects can be 
arranged in M! ways.

*Now, total count of objects is (n-m+1). 
Ex. if, n = 3 and m=2 then if we treat 2 objects as single entity/object we will have 2 (3-2+1=2) objects.

*Since, it is permutation, (n-m+1) can be arranged in (n-m+1)! ways.
i.e. (n-m+1) P (n-m+1). 

*Now; from step 1 and step 3; M objects can be arranged in M! ways, AND (n-m+1) 
objects can be arranged in (n-m+1) ways.
Hence, total number of ways of selection is- Step 1 * Step 3.
Which is , M!*(n-m+1)!
Thanks!!
