# Thread: permutation when p things are always present

1. ## permutation when p things are always present

hi,
i have a problem in understanding a combinatorics formula in my algebra book.
hope i am posting in the right section, i found no section on combinatorics,sorry.
it says that the no of permutations of n different things taken r at a time ,when p particular things are to be always included in each arrengement is
p! (r-p+1) ((n-p) permutation (r-p))
i proceeded in the following way-let me first find the arrengements of the things other than the p ones. that is n-p P r-p.
now the p things can arrenge in p! ways.
but i cannot understand how (r-p+1) is coming here.
thanks in advance for any help.

2. Hello mathlove

Welcome to Math Help Forum!
Originally Posted by mathlove
hi,
i have a problem in understanding a combinatorics formula in my algebra book.
hope i am posting in the right section, i found no section on combinatorics,sorry.
it says that the no of permutations of n different things taken r at a time ,when p particular things are to be always included in each arrengement is
p! (r-p+1) ((n-p) permutation (r-p))
i proceeded in the following way-let me first find the arrengements of the things other than the p ones. that is n-p P r-p.
now the p things can arrenge in p! ways.
but i cannot understand how (r-p+1) is coming here.
thanks in advance for any help.
Are you sure you have copied this formula correctly? I don't see where it comes from either. But neither do I agree with the answer you have suggested!

Let us imagine that we have $\displaystyle r$ empty boxes to fill, and $\displaystyle p$ of them must be filled from a fixed set of $\displaystyle p$ items.

The first of these $\displaystyle p$ items can go in any one of the $\displaystyle r$ boxes; the next in any one of the remaining $\displaystyle (r-1)$ boxes; and so on. The number of ways of filling the first $\displaystyle p$ boxes from the fixed set is therefore
$\displaystyle r(r-1)(r-2)...(r-p+1)={^rP}_p$
There are now $\displaystyle (r-p)$ remaining boxes to be filled with items to be chosen from the remaining $\displaystyle (n-p)$. This can clearly be done in $\displaystyle {^{(n-p)}P}_{(r-p)}$ ways.

So I reckon the total number of ways is
$\displaystyle {^rP}_p\times{^{(n-p)}P}_{(r-p)}$

3. That statement is really difficult to read and understand.

Let me guess. Let’s say we have ten boys among which there are three brothers.
How many ways to from a line of these ten so that the three bothers are together?
The answer is $(3!)(8!)$. There are seven other boys and a group of three, making eight ‘distinct’ items to arrange. But one of those can be further arranged in $3!$ ways.

Is that what this question is about?
If not, then please try to make it clear.

4. Hello everyone

I don't see any particular ambiguity in the way the question is phrased. I have assumed that it means:
How many ways are there of choosing and arranging $\displaystyle r$ items from a set of $\displaystyle n$, if a certain group of $\displaystyle p$ items must always be included in the arrangement?
The question Plato has answered is quite different. He has interpreted it as:
How many ways are there of arranging $\displaystyle n$ items, if a certain group of $\displaystyle p$ items must be placed together in a block within the arrangement?
There doesn't seem any room for the variable $\displaystyle r$ within this latter interpretation.

5. thank you so much,grandad and plato.
yes, actually it means that i always need the p fixed things to remain in each of my arrengement,but not necessarily together.
but i have the formula correctly copied.i have checked the language and the formula again...
but i am having a problem with (r permutation p)which means i am finding the no of arrengements of r objects,taken p at a time. it is giving me a strange feeling that i am not fixing the p things in all arrengements.
say i have the letters a,b,c,d,e,f of which i am asked to find the no of 4 letter words i can form ,where each word must have c and e.(ie. n=6,r=4, ,p=2)
now if i only think about the letters other than c and e, i have 2 places to be filled from 4letters -a,b,d,f =4P2
now to put c and e ,i have 3 blank spaces around the 2 words(other than c and e)chosen already by the above process.
now c has 3 options and d also has 3 options, (because they can sit together),so 3*3*4P2?

6. Originally Posted by mathlove
yes, actually it means that i always need the p fixed things to remain in each of my arrengement,but not necessarily together.

giving me a strange feeling that i am not fixing the p things in all arrangements.
There are $\displaystyle\binom{N-p}{r-p}=\displaystyle\frac{(N-p)!}{(r-p)!(N-r)!}$ to choose the $r$ items among which the specific $p$ items must be.
Then there are $r!$ ways to arrange those.

However, I am given pause when reading you phrase, “not fixing the p things in all arrangements”
What does that mean? Is there a fixed place for the $p$ items?

7. Hello mathlove
Originally Posted by mathlove
thank you so much,grandad and plato.
yes, actually it means that i always need the p fixed things to remain in each of my arrengement,but not necessarily together.
but i have the formula correctly copied.i have checked the language and the formula again...
but i am having a problem with (r permutation p)which means i am finding the no of arrengements of r objects,taken p at a time. it is giving me a strange feeling that i am not fixing the p things in all arrengements.
say i have the letters a,b,c,d,e,f of which i am asked to find the no of 4 letter words i can form ,where each word must have c and e.(ie. n=6,r=4, ,p=2)
now if i only think about the letters other than c and e, i have 2 places to be filled from 4letters -a,b,d,f =4P2
now to put c and e ,i have 3 blank spaces around the 2 words(other than c and e)chosen already by the above process.
now c has 3 options and d also has 3 options, (because they can sit together),so 3*3*4P2?
No. You are assuming here that the other two letters, say a and b are not adjacent to one another.

You must position the letters c and e first, choosing 2 empty 'boxes' from 4, and then arranging the letters within these boxes. This can be done in ${^4P}_2$ ways. The remaining two empty spaces can then be filled from the remaining 4 letters in ${^4P}_2$ ways.

Total: ${^4P}_2\times{^4P}_2$ ways.