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I am trying to find out No of permutations of n different things, taken $r$ at a time, when p particular things is to be always included in each arrangement:

first keep aside $p$ particular thing then the remaining $n-p$ elements can be arranged in $P(n-p,r-p)$ ways, while $p$ particular elements can be arranged in $p!$ ways. So total arrangement is $P(n-p,r-p)\cdot p!$ ways. Also the $r-p$ elements with $p$ particular thing together can be treated as $r-p+1$ elements, whose arrangement should be $(r-p+1)!$. So the final answer should be $P(n-p,r-p)\cdot p!\cdot (r-p+1)!$. I am not very sure about my answer. So can someone justify if am correct or not.

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If $p$ particular things are to be included among the $r$ distinct objects that are being arranged, we must select $r - p$ of the remaining $n - p$ objects, which can be done in $\binom{n - p}{r - p}$ objects. We can then arrange the $r$ objects in $r!$ ways. Hence, there are $$\binom{n - p}{r - p}r!$$ possible permutations that include the $p$ particular objects.

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  • $\begingroup$ Yes, but I guess if my question were p particular things are always included and they will come together then my answer were correct. Thanks for edit and answer. $\endgroup$
    – user525448
    Commented Jan 28, 2018 at 5:18
  • $\begingroup$ If the $p$ particular things are together, there are still $\binom{n - p}{r - p}$ ways of selecting the remaining objects. As you realized, that gives us $r - p + 1$ objects to arrange, the block of $p$ objects and the additional $r - p$ objects. They can be arranged in $(r - p + 1)!$ ways. The $p$ objects within the block can be arranged in $p!$ ways. Thus, in the case, the $p$ particular objects are together, you should obtain $$\binom{n - p}{r - p}(r - p + 1)!p!$$ possible arrangements. $\endgroup$ Commented Jan 28, 2018 at 12:06

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