Question 1121121
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There are two ways to interpret and solve it.


<U>1</U>.

<pre>
There are  {{{C[10]^2}}}  ways to select two children of 10 for vanilla ice cream;

There are  {{{C[8]^3}}}  ways to select three children of remaining 8 for chocolate cones;

There are  {{{C[5]^4}}}  ways to select four children of remaining 5 for strawberry cones;

and there is just NO CHOICE (the unique selection) for this lucky one who will get the pistachio cone.


Now take the product  {{{C[10]^2}}}.{{{C[8]^3}}}.{{{C[5]^4}}}  to get your answer.
</pre>

<U>2.</U>

<pre>
It is the number of all distinguishable permutations of 10 objects, two of which are indistinguishable vanilla ice cream;


the other three are indistinguishable chocolate cones, and the other four are indistinguishable strawberry cones.


In all, there are  {{{10!/(2!*3!*4!)}}} such distinguishable arrangements.
</pre>

Both answers coincide, for your fortune.


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On the last subject, see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/Arranging-elements-of-sets-containing-undistinguishable-elements.lesson>Arranging elements of sets containing indistinguishable elements</A> 

in this site.