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Could someone please explain the difference between these two cases: 1. No. Of ways to arrange n distinguishable balls in r urns. 2. No. Of ways to arrange n balls in r urns when the balls are identical.

It would be great if someone could explain them taking specific values of n and r.

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    The difference between the two cases... take two balls and two urns., the balls are labeled $A$ and $B$, and the urns are represented by brackets and location., $[A][B]$ is considered a different outcome than $[B][A]$. For the balls being identical, these are considered to be the same outcome as they are both $[\star][\star]$. – 2017-02-02
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    @JMoravitz why not post that as an answer? – 2017-02-02
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    @NathanielMayer because someone will inevitably downvote his perfectly fine answer just because the question is ill worded :/ – 2017-02-02
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    @Nathaniel really, if anything it was because that would legitimize this as being a good enough question to properly answer, which I don't think it is. – 2017-02-02
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    Missing from the Question is whether the *urns* are distinguishable or not (i.e. identical urns). – 2017-02-02

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Hint: We should also consider if urns are distinguishable or not. If this is not explicitly stated, we typically assume they are indistinguishable (but should state this assumption in an answer). Here is an example with $n=3$ balls and $r=4$ urns.

We think of $N=\{1,2,3\}$ as a set of balls and of $R=\{a,b,c,d\}$ as a set of urns. A function $f:N\rightarrow R$ is considered as placing each ball into some urn.

We consider four functions $j,k,l,m: N\rightarrow R$ by \begin{array}{lclcllcl} j(1)&=&j(2)&=&a,&\qquad j(3)&=&b\\ k(1)&=&k(3)&=&a,&\qquad k(2)&=&b\\ l(1)&=&l(2)&=&b,&\qquad l(3)&=&d\\ m(2)&=&m(3)&=&b,&\qquad m(1)&=&c\\ \end{array}

Four functions with distinguishable balls and urns:

                                 enter image description here

with balls indistinguishable:

                                 enter image description here

with urns indistinguishable:

                                 enter image description here

with balls and urns indistinguishable:

                                 enter image description here

Note: Some more information is given in this answer.

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In the first case, let's make it simple and consider 2 balls and 2 urns. Have the balls be labeled X and Y, respectfully. Since in the first case the balls are not identical, this means that the probability of choosing X is not equal to choosing Y. The these events are not equally probable, no matter in what urns they are pulled from. In the second case, we have balls X and Y, which are identical. This means that the probability of choosing X then Y is the same as choosing Y then X afterwards. The key thing to remember is the word identical.

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Both the balls and the urns are distinguishable

$n=1,r=1$ gives $1!$ ways

$n=2,r=2$ gives $2!$ ways

$n=6,r=10$ you have a choice of selecting 6 urns from 10 and then put the balls in them that gives you $10\choose6$.$6!$

In case of balls being identical you can disting which ball went in to which urn so if $n=6,r=10$ the answer is just $10\choose 6$

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    $10$ urns, $6$ balls, all distinguishable, there are $10^{6}$ ways to arrange. Balls indistinguishable urns distinguishable there are $\binom{15}{5}$ ways. Balls distinguishable urns indistinguishable there are $\sum\limits_{r=1}^6\left\{\begin{smallmatrix}10\\r\end{smallmatrix}\right\}$ ways... none of which match your given answers. Remember that multiple balls can go to the same urn, you are not limited to just one ball per urn. – 2017-02-02