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I have a question and solution as well ,but I don't understand it: If r distinct objects are to be distributed into n distinct boxes each box can hold any number of objects and the ordering of objects in each box matters,

Then we can put first object in n ways. For placing the $2^{\text nd}$ objects we presume that we have divided the box containing $1^{\text st}$ object into 2 parts (one part left to the object and the other right to the object). So $2^{\text nd}$ object can be placed in n+1 ways. Continuing like this ,we can place $i^{\text th}$ object in $n+i-1$ ways.

total no. of ways for placing r objects are : $n(n+1)\ldots(n+r-1)$.I can't understand why is it $n+1$ ways for $2^{\text nd}$ object?

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Because you can have $n-1$ empty boxes to put it in, and then two different ways of putting it in the box that already has the first object in, either to the left or to the right, as you put it.

One way of thinking about it might be packing dinner plates into cardboard boxes. Say you have 10 plates and 3 boxes, you can put the first plate into any of the 3 boxes. The second plate you can put in any of the 2 empty boxes, or you can put it on top of the first plate, or underneath it. Hence there are 4 ways to pack the second plate.

Hope this helps.

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