In this theorem I understand that if there are n objects and r number of objects are taken at a time and if any one object is always included in any arrangement then what we do is ($n-1!/r-1!$)-(i) but why in the formula are we multiplying (i) by r?
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Because the order matters.
After we choose $r-1$ of them out of the remaining $n-1$ iterms. We first sort the order.
After which, we still need to decide a position for the special object, there are $r$ options for the position of the object, hence by multiplication principle, we multiply by $r$.
Siong Thye Goh
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okay, so its like we first find out all the arrangements with the n-1 objects at remaining r-1 positions and then we multiply by r so that we can get all the arrangements in which the special object is placed as in permutation the position matters and that object can be placed in r ways. – Mad Dawg Dec 30 '18 at 06:07