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) !$

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!!
