5
$\begingroup$

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:

  1. First we have choose r out of n: $^nC_r=\frac{n!}{(n-r)!r!}$
  2. Then choose m out of r: $^rC_m=\frac{r!}{(r-m)!m!}$
  3. Next arrange m elements : $m! (r-m+1)!$
  4. 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) !$

$\endgroup$
2
  • $\begingroup$ The text underlined in beige is fishy. A variable $r$is introduced in the formulation of the problem, but it does not appear in the result. What do you mean by "taken $r$ at a time"? $\endgroup$ Jun 2, 2015 at 19:24
  • $\begingroup$ Yep I too noticed that. But you can find many sources online stating this formula. For example, this and this $\endgroup$
    – Mahesha999
    Jun 2, 2015 at 19:54

3 Answers 3

0
$\begingroup$

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)

$\endgroup$
2
  • $\begingroup$ You have $n-m+1$ elements to order. nope I guess. Out of r I have m elements together, so its $r-m+1$ $\endgroup$
    – Mahesha999
    Jun 2, 2015 at 14:47
  • $\begingroup$ Also check the answer given here by cheeser1. He/she also reaches till where I reached, no further. $\endgroup$
    – Mahesha999
    Jun 2, 2015 at 14:48
0
$\begingroup$

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:

  1. Select the m objects: - assuming they are specified, there is only one way to do this.
  2. Select the rest of the r objects to be arranged: (n-m)!/((r-m)!(n-m+r)!) ways to do this
  3. arrange the m specified objects: m! ways - this forms a single large object
  4. 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.

$\endgroup$
0
$\begingroup$

This is permutation question, so objects can come in different order.

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

  2. 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.

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

  4. 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!!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .